WEBVTT 00:00:17.679 --> 00:00:23.420 This video is about the exponent rules, rules that govern expressions like two to the fifth, 00:00:23.420 --> 00:00:26.040 or x to the n. 00:00:26.039 --> 00:00:32.839 two to the fifth is just shorthand for two times two times two times two times two, written 00:00:32.840 --> 00:00:40.430 five times. And similarly x to the n is just x multiplied by itself. And times 00:00:40.429 --> 00:00:47.390 when we write these expressions, the number on the bottom that's being multiplied by itself 00:00:47.390 --> 00:00:54.000 is called the base. And the number at the top telling us how many times we're multiplying 00:00:54.000 --> 00:00:58.280 the base by itself is called the exponent. 00:00:58.280 --> 00:01:02.969 Sometimes the exponent is also called the power. 00:01:02.969 --> 00:01:10.650 The product rule says that a 5x to the power of n times x to the power of m, that's the 00:01:10.650 --> 00:01:19.860 same thing as x to the n plus m power. In other words, when I multiply two expressions 00:01:19.859 --> 00:01:23.060 with the same base, 00:01:23.060 --> 00:01:27.040 then I can add their exponents. 00:01:27.040 --> 00:01:33.180 For example, if I have two cubed times two to the fourth, that's equal to two to the 00:01:33.180 --> 00:01:40.060 seventh. And that makes sense, because two cubed times two to the fourth, means I multiply 00:01:40.060 --> 00:01:46.710 two by itself three times. And then I multiply that by two multiplied by itself four times. 00:01:46.709 --> 00:01:53.849 And in the end, I have to multiplied by itself seven times, 00:01:53.849 --> 00:01:56.629 which is two to the seventh, 00:01:56.629 --> 00:02:01.180 I'm just adding up the number of times as multiplied in each piece, to get the number 00:02:01.180 --> 00:02:04.228 of times as multiplied total. 00:02:04.228 --> 00:02:10.689 The quotient rule says that if I have x to the n power divided by x to the m power, that's 00:02:10.689 --> 00:02:17.750 equal to x to the n minus m power. In other words, if I divide two expressions with the 00:02:17.750 --> 00:02:19.219 same base, 00:02:19.219 --> 00:02:23.609 then I can subtract their exponents. 00:02:23.610 --> 00:02:31.340 For example, three to the six divided by three squared is going to be three to the six minus 00:02:31.340 --> 00:02:37.939 two, or three to the fourth. And this makes sense, because three to the six means I multiply 00:02:37.939 --> 00:02:45.270 three by itself six times, and then I divide that by three multiplied by itself twice. 00:02:45.270 --> 00:02:50.120 So when I cancel out threes, I have four threes left, 00:02:50.120 --> 00:02:54.430 notice that I have to subtract the number of threes on the bottom, from the number of 00:02:54.430 --> 00:03:01.900 threes at the top to get my number of threes remaining. That's why subtract my exponents. 00:03:01.900 --> 00:03:08.120 The power rule tells us if I have x to the n power raised to the m power, that's the 00:03:08.120 --> 00:03:17.980 same thing as x to the n times M power. In other words, when I raise a power to a power, 00:03:17.979 --> 00:03:21.530 I get to multiply the exponents. 00:03:21.530 --> 00:03:29.729 For example, five to the fourth cubed is equal to five to the four times three, or five to 00:03:29.729 --> 00:03:34.859 the 12th. And this makes sense, because five to the fourth cubed can be thought of as five 00:03:34.860 --> 00:03:40.610 to the fourth times five to the fourth, times five to the fourth, expanding this out some 00:03:40.610 --> 00:03:49.670 more, that's five times five times five times five times the same thing times the same thing 00:03:49.669 --> 00:03:51.918 again. 00:03:51.919 --> 00:04:03.620 So I have three groups of four, or five, which is a total of three times four, or 12. fives. 00:04:03.620 --> 00:04:09.239 The next rule involves what happens when I raise a number, or a variable to the zeroeth 00:04:09.239 --> 00:04:14.299 power, it turns out that anything to the zeroeth power is equal to one. 00:04:14.299 --> 00:04:20.930 Usually, this is just taken as a definition. But here's why it makes sense to me. 00:04:20.930 --> 00:04:27.180 If you have something like two cubed divided by two cubed, Well, certainly that has to 00:04:27.180 --> 00:04:35.109 equal one, anything divided by itself is just one. But using the quotient rule, 00:04:35.110 --> 00:04:40.669 we know that this is the same thing as two to the three minus three, because when we 00:04:40.668 --> 00:04:47.659 divide two things with the same base, we get to subtract their exponents. Therefore, this 00:04:47.660 --> 00:04:53.849 is the same thing as two to the zero. So two to the zero has to equal one in order to make 00:04:53.848 --> 00:04:55.879 it work with the quotient rule. 00:04:55.879 --> 00:05:00.089 And the same argument shows that anything to the zero power 00:05:00.089 --> 00:05:02.948 has to be equal to one. 00:05:02.949 --> 00:05:07.770 What happens when we take something to a negative power, 00:05:07.769 --> 00:05:15.799 x to the n is equal to one over x to the n. Again, most people just take this as a definition 00:05:15.800 --> 00:05:20.310 of a negative exponent. But here's why it makes sense. 00:05:20.310 --> 00:05:26.649 If I take something like five to seven times five to the negative seven, then buy the product 00:05:26.649 --> 00:05:33.429 roll the SAS to equal five to the seven plus negative seven, which is five to the zero, 00:05:33.430 --> 00:05:35.680 and we just said that that is equal to one. 00:05:35.680 --> 00:05:41.209 Now I have the equation of five to the seventh times five to the negative seventh equals 00:05:41.209 --> 00:05:47.750 one, if I divide both sides by five, the seventh, I get that five to the negative seventh has 00:05:47.750 --> 00:05:52.759 to equal one over five to the seventh. So that's where this rule about negative exponents 00:05:52.759 --> 00:05:58.009 comes from. That has to be true in order to be consistent with the product rule. 00:05:58.009 --> 00:06:03.509 Finally, let's look at a fractional exponents. What does an expression like x to the one 00:06:03.509 --> 00:06:14.479 over N really mean? Well, it means the nth root of x, for example, 64 to the 1/3 power 00:06:14.478 --> 00:06:20.610 means the cube root of 64, which happens to be four, and nine to the one half means the 00:06:20.610 --> 00:06:26.120 square root of nine, which is usually written without that little superscript up there. 00:06:26.120 --> 00:06:29.959 Now, the square root of nine is just three. 00:06:29.959 --> 00:06:39.138 fractional exponents also makes sense. For example, if I have five to the 1/3, and I 00:06:39.139 --> 00:06:46.970 cube that, then by the power role, that's equal to five to the 1/3, times three, 00:06:46.970 --> 00:06:55.150 which is just five to the one or five. So in other words, five to the 1/3, is the number 00:06:55.149 --> 00:07:01.388 that when you cube it, you get five. And that's exactly what's meant by the cube root of five, 00:07:01.389 --> 00:07:07.509 the cube root of five is also a number that when you cube it, you get five. 00:07:07.509 --> 00:07:12.660 The next rule tells us we can distribute an exponent over a product. 00:07:12.660 --> 00:07:18.939 In other words, if we have a product, x times y, all raised to the nth power, that's equal 00:07:18.939 --> 00:07:22.749 to x to the n times y to the N. 00:07:22.749 --> 00:07:27.219 For example, five times seven, 00:07:27.218 --> 00:07:33.098 all raised to the third power is equal to five cubed times seven cubed. And this makes 00:07:33.098 --> 00:07:39.800 sense, because five times seven, all raised to the cube power can be expanded as five 00:07:39.800 --> 00:07:46.520 times seven, times five times seven, times five times seven. But if I rearrange the order 00:07:46.519 --> 00:07:51.740 of multiplication, this is the same thing as five times five times five, times seven 00:07:51.740 --> 00:07:56.668 times seven times seven, or five cubed times seven cubed. 00:07:56.668 --> 00:08:02.248 Similarly, we could distribute an exponent over a quotient, if we have the quotient X 00:08:02.249 --> 00:08:09.889 over Y, all raised to the n power, that's the same as x to the n over y to the N. For 00:08:09.889 --> 00:08:15.840 example, two sevenths raised to the fifth power is the same thing as two to the fifth 00:08:15.839 --> 00:08:23.038 over seven to the fifth. This makes sense, because two sevens to the fifth can be expanded 00:08:23.038 --> 00:08:31.079 as two sevens multiply by itself five times, which can be written rewritten as two multiplied 00:08:31.079 --> 00:08:37.649 by itself five times divided by seven multiplied by itself five times, and that's two to the 00:08:37.649 --> 00:08:43.348 fifth, over seven to the fifth, as wanted. 00:08:43.349 --> 00:08:47.680 We've seen that we can distribute an exponent 00:08:47.679 --> 00:08:51.039 over multiplication, and division. 00:08:51.039 --> 00:08:56.969 But be careful, because we cannot distribute next bowknot. 00:08:56.970 --> 00:09:06.600 over addition, or subtraction, for example, a plus b to the n is not generally equal to 00:09:06.600 --> 00:09:14.009 a to the n plus b to the n, a minus b to the n is not generally equal to a to the n minus 00:09:14.009 --> 00:09:19.448 b to the n. And if you're not sure, just try an example with numbers. 00:09:19.448 --> 00:09:28.609 For example, two plus three squared is not the same thing as two squared plus three squared, 00:09:28.610 --> 00:09:36.430 and two minus three squared is definitely not equal to two squared minus three squared. 00:09:36.429 --> 00:09:42.059 In this video, I gave eight exponent rules, which I'll list again here. There's the product 00:09:42.059 --> 00:09:43.539 rule, 00:09:43.539 --> 00:09:46.559 the quotient rule, the power rule, 00:09:46.559 --> 00:09:55.629 the zero exponent, the negative exponent, the fractional exponent, 00:09:55.629 --> 00:10:00.009 and the two rules involving distributing exponents. 00:10:00.009 --> 00:10:02.370 Across multiplication, 00:10:02.370 --> 00:10:06.350 and division. 00:10:06.350 --> 00:10:11.540 In another video, I'll use these exponent rules to rewrite and simplify expressions 00:10:11.539 --> 00:10:13.649 involving exponents. 00:10:13.649 --> 00:10:20.309 In this video, I'll work out some examples of simplifying expressions using exponent 00:10:20.309 --> 00:10:21.359 rules. 00:10:21.360 --> 00:10:22.740 I'll start by reviewing the exponent rules. 00:10:22.740 --> 00:10:30.690 The product rule says that when you multiply two expressions with the same base, you add 00:10:30.690 --> 00:10:36.320 the exponents. The quotient rule says that when you divide two expressions with the same 00:10:36.320 --> 00:10:42.790 base, you subtract the exponents. The power rule says that when you take a power to a 00:10:42.789 --> 00:10:50.539 power, you multiply the exponents. The power of zero rule says that anything to the zero 00:10:50.539 --> 00:10:54.740 power is one, as long as the base is not zero. 00:10:54.740 --> 00:10:59.589 Since zero to the zero is undefined, it doesn't make sense. 00:10:59.589 --> 00:11:07.600 negative exponents to evaluate x to the minus n, we take the reciprocal one over x to the 00:11:07.600 --> 00:11:16.250 n. To evaluate a fractional exponent, like x to the one over n, we take the nth root 00:11:16.250 --> 00:11:17.278 of x, 00:11:17.278 --> 00:11:24.669 we can distribute an exponent over a product, a times b to the n is equal to A to the N 00:11:24.669 --> 00:11:31.159 times b to the n. And we can distribute an exponent over a quotient a over b to the n 00:11:31.159 --> 00:11:36.198 is a to the n over b to the n. 00:11:36.198 --> 00:11:40.329 In the rest of this video, we'll use these exponent rules to simplify expressions. 00:11:40.330 --> 00:11:47.399 For our first example, we want to simplify three times x to the minus two divided by 00:11:47.399 --> 00:11:51.769 x to the fourth, there's several possible ways to proceed. 00:11:51.769 --> 00:12:00.320 For example, we could use the negative exponent rule dr x to the minus two as one over x squared, 00:12:00.320 --> 00:12:03.589 all that gets divided by x to the fourth, still, 00:12:03.589 --> 00:12:09.860 notice that we only take the reciprocal of the x squared, the three stays where it is. 00:12:09.860 --> 00:12:16.818 And that's because the exponent of negative two only applies to the x not to the three. 00:12:16.818 --> 00:12:23.870 Now if we think of three as three over one, we have a product of two fractions and our 00:12:23.870 --> 00:12:30.600 numerator. And so we evaluate that by taking the product of the numerators times the product 00:12:30.600 --> 00:12:36.709 of the denominators, which is three over x squared, all divided by x to the fourth, 00:12:36.708 --> 00:12:45.690 I can think of x to the fourth as x to the fourth over one. So now I have a fraction 00:12:45.691 --> 00:12:52.980 of a fraction, which I can evaluate by multiplying by the reciprocal, that simplifies to three 00:12:52.980 --> 00:13:00.019 times one divided by x squared times x to the fourth, which is three over x to the six, 00:13:00.019 --> 00:13:07.959 using the product rule. Since x squared times x to the fourth is equal to x to the two plus 00:13:07.958 --> 00:13:12.698 four, or x to the six. 00:13:12.698 --> 00:13:20.599 An alternate way of solving this problem 00:13:20.600 --> 00:13:21.600 is to start by using the quotient rule, 00:13:21.600 --> 00:13:25.590 I can rewrite this as three times x to the minus two over x to the fourth, and by the 00:13:25.590 --> 00:13:31.560 quotient rule, that's three times x to the minus two minus four, or three times x to 00:13:31.559 --> 00:13:32.849 the minus six. 00:13:32.850 --> 00:13:40.960 Now using the negative exponent roll, x to the minus six is one over x to the six. And 00:13:40.960 --> 00:13:49.600 this product of fractions simplifies to three over x to the six, the same answer I got before. 00:13:49.600 --> 00:14:00.240 The second problem can be solved in similar ways. Please pause the video and try it before 00:14:00.240 --> 00:14:01.490 going on. 00:14:01.490 --> 00:14:04.568 One way to simplify would be to use the negative exponent rule first, and rewrite y to the 00:14:04.568 --> 00:14:08.299 minus five as one over wide the fifth. 00:14:08.299 --> 00:14:18.240 thinking of this as a fraction, divided by a fraction, I can multiply by the reciprocal 00:14:18.240 --> 00:14:28.549 and get four y cubed y to the fifth over one by the product rule. The numerator here is 00:14:28.549 --> 00:14:35.528 four y to the eight. And so my final answer is just for one of the eight. 00:14:35.528 --> 00:14:41.720 Alternatively, I could decide to use the quotient rule first. 00:14:41.720 --> 00:14:48.320 As in the previous problem, I can write this as for y cubed minus negative five by the 00:14:48.320 --> 00:14:53.670 quotient rule. And so that's for y to the eighth as before. 00:14:53.669 --> 00:14:59.319 I'd like to show you one more method to solve these two problems, kind of a shortcut method 00:14:59.320 --> 00:15:00.320 before we 00:15:00.320 --> 00:15:08.300 To go on, that shortcut relies on the principle that a negative exponent in the numerator 00:15:08.299 --> 00:15:14.269 corresponds to a positive exponent in the denominator. For example, the x to the negative 00:15:14.269 --> 00:15:22.620 two in the numerator here, after some manipulations became an X to the positive two in the denominator. 00:15:22.620 --> 00:15:27.950 Furthermore, a negative exponent in the denominator 00:15:27.950 --> 00:15:32.769 is equivalent to a positive exponent in the numerator. 00:15:32.769 --> 00:15:39.360 That's what happened when we had the y to the negative five in the denominator, and 00:15:39.360 --> 00:15:42.829 translated into a y to the positive five in the numerator. 00:15:42.828 --> 00:15:52.198 Sometimes people like to talk about this principle, by saying that you can pass a factor 00:15:52.198 --> 00:15:54.049 across the fraction bar 00:15:54.049 --> 00:16:04.659 by switching the sine of the exponent that is making a positive exponent negative, or 00:16:04.659 --> 00:16:07.730 a negative exponent positive. 00:16:07.730 --> 00:16:11.230 Let's see how this principle gives us a shortcut for solving these two problems. 00:16:11.230 --> 00:16:20.819 In the first problem, 3x to the minus two over x to the four, we can move the negative 00:16:20.818 --> 00:16:25.809 exponent in the numerator and make it a positive exponent the denominator, so we get three 00:16:25.809 --> 00:16:32.708 over x to the four plus two or x to the six. 00:16:32.708 --> 00:16:40.198 In the second example, for y cubed over y to the minus five, we can change the Y to 00:16:40.198 --> 00:16:45.958 the minus five in the denominator into a y to the five in the numerator and get our final 00:16:45.958 --> 00:16:51.578 answer of four y to the three plus five or eight. 00:16:51.578 --> 00:16:54.489 We'll use this principle again in the next problems. 00:16:54.490 --> 00:17:01.490 In this example, notice that I have I have y's in the numerator and the denominator, 00:17:01.490 --> 00:17:09.819 and also Z's in the numerator and the denominator. In order to simplify, I'm going to try to 00:17:09.819 --> 00:17:13.389 get all my y's either in the numerator or the denominator. And similarly for the Z's. 00:17:13.390 --> 00:17:19.630 Since I have more y's in the denominator, let me move this y to the three downstairs 00:17:19.630 --> 00:17:26.640 and make it a y to the negative three. I'm using the principle here that a positive exponent 00:17:26.640 --> 00:17:31.820 in the numerator corresponds to a negative exponent in the denominator. 00:17:31.819 --> 00:17:37.439 Now since I have a positive exponent, z in the numerator and a negative exponent denominator, 00:17:37.440 --> 00:17:44.279 and I want to get rid of negative exponents, I'm going to pass the Z's to the numerator 00:17:44.279 --> 00:17:51.289 as e to the minus two in the denominator, it comes as e to the plus two in the numerator. 00:17:51.289 --> 00:17:57.649 Notice that my number seven doesn't move. And when I do any of these manipulations, 00:17:57.650 --> 00:17:59.820 because it doesn't have an exponent, and the exponent of negative two, for example, only 00:17:59.819 --> 00:18:01.619 applies to the Z not to the seven. 00:18:01.619 --> 00:18:10.929 Now that I've got all my Z's in the numerator and all my y's in the denominator, it's easy 00:18:10.930 --> 00:18:13.070 to clean this up using the product rule. 00:18:13.069 --> 00:18:16.750 And I have my simplified expression. 00:18:16.750 --> 00:18:22.150 In this last example, we have a complicated expression raised to a fractional power. 00:18:22.150 --> 00:18:26.750 I'm going to start by simplifying the expression inside the parentheses. 00:18:26.750 --> 00:18:34.329 I can bring all my y's downstairs and all my x's upstairs and get rid of negative exponents 00:18:34.329 --> 00:18:35.809 at the same time. 00:18:35.809 --> 00:18:40.139 In other words, I can rewrite this as 25x to the fourth, 00:18:40.140 --> 00:18:48.290 I'll bring the Y to the minus five downstairs and make it y to the fifth on the denominator, 00:18:48.289 --> 00:18:54.339 bring the x to the minus six upstairs and make it x to the sixth the numerator, and 00:18:54.339 --> 00:19:00.199 then I still have the Y cubed on the denominator, all that's raised to the three halves power. 00:19:00.200 --> 00:19:07.130 Using the product rule, I can rewrite the expression on the inside of the parentheses 00:19:07.130 --> 00:19:12.120 as 25x to the 10th over y to the eighth. 00:19:12.119 --> 00:19:21.699 Recall that we're allowed to distribute an exponent across a product or across a quotient. 00:19:21.700 --> 00:19:27.710 When I distribute my three halves power, 00:19:27.710 --> 00:19:36.789 I get 25 to the three halves times x to the 10th to the three halves divided by y to the 00:19:36.789 --> 00:19:38.950 eighth to the three halves. 00:19:38.950 --> 00:19:46.830 Now the power rule tells me when I have a power to a power, I get to multiply the exponents. 00:19:46.829 --> 00:19:53.789 So I can rewrite this as 25 to the three halves times x to the 10 times three halves of our 00:19:53.789 --> 00:20:01.039 y to the eight times three halves. In other words, 25 to the three halves times x to the 00:20:01.039 --> 00:20:10.289 15th over y to the 12th. Finally, I need to rewrite 25 to the three halves. Since three 00:20:10.289 --> 00:20:17.059 halves can be thought of as three times one half, or as one half times three, I can write 00:20:17.059 --> 00:20:27.019 25 to the three halves as 25 to the three times one half, or as 25 to the one half times 00:20:27.019 --> 00:20:28.019 three. 00:20:28.019 --> 00:20:33.299 Well, using the power rule in reverse, I can think of this as 25 cubed to the one half, 00:20:33.299 --> 00:20:39.710 or as 25 to the one half cubed. Since when I take a power to a power, I multiply the 00:20:39.710 --> 00:20:40.710 exponents 00:20:40.710 --> 00:20:47.819 25 cubed to the one half might be hard to evaluate, since 25 cubed is a huge number, 00:20:47.819 --> 00:20:58.439 but 25 to the one half is just the square root of 25. So I have the square root of 25 00:20:58.440 --> 00:21:02.100 cubed, or five cubed, which is 125. 00:21:02.099 --> 00:21:09.139 Therefore, my original expression is going to be 120 5x to the 15 over y to the 12th. 00:21:09.140 --> 00:21:18.400 In this video, we use the exponent rules to simplify complicated expressions. 00:21:18.400 --> 00:21:23.800 This video goes through a few tricks for simplifying expressions with radicals in them. 00:21:23.799 --> 00:21:30.669 Recall that this notation means the nth root of x, so this notation here means the cube 00:21:30.670 --> 00:21:38.360 root of eight, the number that when you cube it, you get eight, that number would be two, 00:21:38.359 --> 00:21:44.079 when we write the root sign without a little number, that just means the square root so 00:21:44.079 --> 00:21:47.230 the two is implied. 00:21:47.230 --> 00:21:54.599 In this case, the square root of 25 is five since five squared is 25. 00:21:54.599 --> 00:22:00.149 Let's start by reviewing some rules for radical expressions. First, if we have the radical 00:22:00.150 --> 00:22:06.940 of a product, we can rewrite that as the product of two radicals. 00:22:06.940 --> 00:22:14.830 For example, the square root of nine times 16 is the same thing as the square root of 00:22:14.829 --> 00:22:21.179 nine times the square root of 16, you can check that both of these evaluate to 12. 00:22:21.180 --> 00:22:29.150 Similarly, it's possible to distribute a radical sine across division, the radical of a divided 00:22:29.150 --> 00:22:37.350 by b is the same thing as the radical of A divided by the radical of B. For example, 00:22:37.349 --> 00:22:45.299 the cube root of 64 over eight is the same thing as the cube root of 64 over the cube 00:22:45.299 --> 00:22:48.980 root of eight, and you can check that both of these evaluated to 00:22:48.980 --> 00:22:57.039 you have to be a little bit careful though, because it's not okay to distribute a radical 00:22:57.039 --> 00:23:03.819 sign across addition. In general, the nth root of a plus b is not equal to the nth root 00:23:03.819 --> 00:23:11.179 of A plus the nth root of b. And similarly, it's not okay to distribute a radical across 00:23:11.180 --> 00:23:14.130 subtraction. 00:23:14.130 --> 00:23:19.450 If you're ever in doubt, you can always check with simple examples. For example, the square 00:23:19.450 --> 00:23:24.490 root of one plus one is not the same thing as the square root of one plus the square 00:23:24.490 --> 00:23:26.450 root of one, 00:23:26.450 --> 00:23:32.269 the right side evaluates to one plus one or two, and the left side is square root of two 00:23:32.269 --> 00:23:34.650 and the irrational number. 00:23:34.650 --> 00:23:40.019 The second expression to show that that fails, I don't think it'll work to use the square 00:23:40.019 --> 00:23:44.889 root of one minus one it'll actually hold in that case, but I can show it's false by 00:23:44.890 --> 00:23:49.890 using say the square root of two minus one, which does not equal the square root of two 00:23:49.890 --> 00:23:53.050 minus the square root of one. 00:23:53.049 --> 00:23:57.240 You might notice that these Rules for Radicals, the ones that hold and the ones that don't 00:23:57.240 --> 00:24:04.420 hold, remind you of rules for exponents. And that's no coincidence. Because radicals can 00:24:04.420 --> 00:24:09.920 be written in terms of exponents. For example, if we look at the first rule, we can rewrite 00:24:09.920 --> 00:24:19.000 this, the nth root of A times B is the same thing as the one of our nth power. And by 00:24:19.000 --> 00:24:26.099 exponent rules, I can distribute an exponent across multiplication. And so this radical 00:24:26.099 --> 00:24:34.819 rule can be restated completely in terms of an exponent rule. Similarly, the second rule 00:24:34.819 --> 00:24:41.119 can be restated in terms of exponents as a or b to the one over n is equal to A to the 00:24:41.119 --> 00:24:46.779 one over n divided by b to the one over n. We can use the relationship between radicals 00:24:46.779 --> 00:24:54.700 and the exponents to rewrite a to the m over n. a to the m over N is the same thing as 00:24:54.700 --> 00:25:00.250 a to the m with the N through taken. That's also the same thing 00:25:00.250 --> 00:25:06.640 As the nth root of A, all taken to the nth power to see whether that's true, think about 00:25:06.640 --> 00:25:13.011 exponent rules. So a to the m over N is the same thing as a to the m, taken to the one 00:25:13.010 --> 00:25:19.029 over nth power. That's because when we take a power to a power, we multiply exponents, 00:25:19.029 --> 00:25:25.680 and M times one over n is equal to M over n. 00:25:25.680 --> 00:25:31.660 But a one over nth power is the same thing as an nth root. And therefore, this expression 00:25:31.660 --> 00:25:39.070 is the same thing as this expression. And that proves the first equivalence. The second 00:25:39.069 --> 00:25:48.169 equivalence can we prove similarly, by writing a to the m over n as a to the one over n times 00:25:48.170 --> 00:25:55.380 M. Again, this is works because when I take the power to the power, I'm multiply exponents, 00:25:55.380 --> 00:26:02.330 one over n times M is the same thing as M over n. But now, these two expressions are 00:26:02.329 --> 00:26:08.119 the same, because the one over nth power is the same as the nth root. 00:26:08.119 --> 00:26:15.729 One mnemonic for remembering these relationships is flower over root. So flour is like power, 00:26:15.730 --> 00:26:20.779 and root is like root, so that tells us we can write a fractional exponent, the M becomes 00:26:20.779 --> 00:26:26.259 the power, and the n becomes the root in either of these two orders. 00:26:26.259 --> 00:26:33.109 Now let's use these rules in some examples. If we want to compute 25 to the negative three 00:26:33.109 --> 00:26:39.159 halves power, well, first I'll use my exponent rule to rewrite that negative exponent as 00:26:39.160 --> 00:26:43.029 one over 25 to the three halves power. 00:26:43.029 --> 00:26:52.839 Next, I'll use the power of a root mnemonic to rewrite this as 25 to the third power square 00:26:52.839 --> 00:26:59.500 rooted, or, as 25 square rooted to the third power. 00:26:59.500 --> 00:27:04.650 I wrote the two's there for the square root for emphasis, but most of the time, people 00:27:04.650 --> 00:27:09.900 will omit this and just write the square root without a little number there. 00:27:09.900 --> 00:27:15.750 Now, I could use either of these two equivalent expressions to continue, but I'd rather use 00:27:15.750 --> 00:27:20.569 this one because it's easier to compute without a calculator. The square root of 25 is just 00:27:20.569 --> 00:27:28.589 five, five cubed is 125. So my answer is one over 125. 00:27:28.589 --> 00:27:34.750 If I tried to compute the cube of 25, first, I'd get a huge number. In general, it's usually 00:27:34.750 --> 00:27:41.859 easier to compute the route before the power when you're working without a calculator. 00:27:41.859 --> 00:27:46.209 Now let's do an example simplifying a more complicated expression with with exponent 00:27:46.210 --> 00:27:52.450 cynet. I want to take the square root of all this stuff. And since I don't really like 00:27:52.450 --> 00:27:58.471 negative exponents, I'm first going to rewrite this as the square root of 60x squared y to 00:27:58.471 --> 00:28:04.319 the sixth over z to the 11th. So I'll change that negative exponent to a positive exponent 00:28:04.319 --> 00:28:07.879 by moving this, this factor to the denominator. 00:28:07.880 --> 00:28:13.450 Now, when you're asked to simplify radical expression, that generally means to pull as 00:28:13.450 --> 00:28:17.950 much as possible, out of the radical side. 00:28:17.950 --> 00:28:23.000 To pull things out of the square root side, I'm going to factor my numbers and try to 00:28:23.000 --> 00:28:28.410 rewrite everything in terms of squares as much as possible. Since the square root of 00:28:28.410 --> 00:28:34.890 a square, those two operations undo each other. So I'll show you what I mean first, our factor 00:28:34.890 --> 00:28:43.210 60. So 60 is going to be two squared times three times five. And I'll just copy everything 00:28:43.210 --> 00:28:47.259 over for now. 00:28:47.259 --> 00:28:52.690 Now I'll break things up into squares as much as possible. So I've got a two squared, and 00:28:52.690 --> 00:28:57.360 three times five, I've already got an x squared, I write the wider the six as y squared times 00:28:57.359 --> 00:29:03.579 y squared times y squared. And I'll write the Z to the 11th as z squared times z squared, 00:29:03.579 --> 00:29:12.048 I guess five times 12345 times when extra z, that should add up to z to the 11th. I 00:29:12.048 --> 00:29:15.410 add all those exponents together. 00:29:15.410 --> 00:29:20.430 Now I know that I can distribute my radical sign across multiplication and division. So 00:29:20.430 --> 00:29:25.860 I'll write this with a zillion different radicals here. 00:29:25.859 --> 00:29:32.479 And every time I see the square root of something squared, I can just cancel those square roots 00:29:32.480 --> 00:29:37.410 and the squares out and get what's what's left here. So So after doing that cancellation, 00:29:37.410 --> 00:29:46.080 I get two times the square root of three times five times x times y times y times y over 00:29:46.079 --> 00:29:53.569 z times itself, I guess five times times the square root of z. And now I can clean that 00:29:53.569 --> 00:29:59.069 up with exponents. I'll write that as the square root of 15 I guess two times a squared 00:29:59.069 --> 00:30:00.189 of 15 00:30:00.190 --> 00:30:07.950 times x times y cubed over z to the fifth the square root of z. 00:30:07.950 --> 00:30:15.480 I'm gonna leave this example as is. But sometimes people prefer to rewrite radical expressions 00:30:15.480 --> 00:30:20.769 without radical signs in the denominator. That's called rationalizing the denominator. 00:30:20.769 --> 00:30:24.789 I won't do it here, but I'll show you how to do it in the next example. 00:30:24.789 --> 00:30:31.859 This example asks us to rationalize the denominator, that means to rewrite as an equivalent expression 00:30:31.859 --> 00:30:36.549 without radical signs in the denominator. 00:30:36.549 --> 00:30:44.069 To get rid of the radical sine and the denominator, I want to multiply my denominator by square 00:30:44.069 --> 00:30:49.389 root of x. But I can't just multiply the denominator willy nilly by something unless I multiply 00:30:49.390 --> 00:30:55.240 the numerator by the same thing. So then just multiply my expression by one in a fancy forum 00:30:55.240 --> 00:30:57.529 and I don't change the value of my expression. 00:30:57.529 --> 00:31:06.319 Now, if I just multiply together numerators 3x squared of x and multiply denominators 00:31:06.319 --> 00:31:12.019 squared of x 10 squared of x is the square root of x squared squared of x squared is 00:31:12.019 --> 00:31:14.210 just x. 00:31:14.210 --> 00:31:19.620 Now I can cancel my access from the numerator denominator, and my final answer is three 00:31:19.619 --> 00:31:24.750 times the square root of x. I rationalize my denominator, and the process got a nicer 00:31:24.750 --> 00:31:26.400 looking expression. 00:31:26.400 --> 00:31:34.110 In this video, we went over the Rules for Radicals. And we simplified some radical expressions 00:31:34.109 --> 00:31:37.889 by working with fractional exponents, 00:31:37.890 --> 00:31:41.360 pulling things out of the radical sign 00:31:41.359 --> 00:31:46.159 and rationalizing the denominator. 00:31:46.160 --> 00:31:51.170 This video goes over some common methods of factoring. Recall that factoring an expression 00:31:51.170 --> 00:31:56.890 means to write it as a product. So we could factor the number 30, by writing it as six 00:31:56.890 --> 00:32:04.670 times five, we could factor it more completely by writing it as two times three times five. 00:32:04.670 --> 00:32:12.670 As another example, we could factor the expression x squared plus 5x plus six by writing it as 00:32:12.670 --> 00:32:16.500 x plus two times x plus three. 00:32:16.500 --> 00:32:23.700 In this video, I'll go over how I get from here to here, how I know how to do the factoring. 00:32:23.700 --> 00:32:29.340 But for right now, I just want to review how I can go backwards how I can check that the 00:32:29.339 --> 00:32:35.259 factoring is correct. And that's just by multiplying out or distributing. If I distribute x plus 00:32:35.259 --> 00:32:40.799 two times x plus three, then I multiply x by x, that gives me x squared, x times three 00:32:40.799 --> 00:32:48.119 gives me 3x. Two times x gives me 2x. And two times three gives me six. So that simplifies 00:32:48.119 --> 00:32:54.569 to x squared plus 5x plus six, which checks out with what I started with. So you can think 00:32:54.569 --> 00:33:00.240 of factoring as the opposite of distributing out. And you can always check your factoring 00:33:00.240 --> 00:33:01.730 by distributing or multiplying out 00:33:01.730 --> 00:33:09.140 a bit of terminology, when I think of an expression as a sum of a bunch of things, then the things 00:33:09.140 --> 00:33:17.610 I sum up are called the terms. But if I think of the same expression as a product of things, 00:33:17.609 --> 00:33:22.929 then the things that I multiply together are called factors. 00:33:22.930 --> 00:33:28.350 Now let's get started on techniques of factoring. When I have to factor something, I always 00:33:28.349 --> 00:33:33.339 like to start by pulling out the greatest common factor, the greatest common factor 00:33:33.339 --> 00:33:38.269 means the largest thing that divides each of the terms. 00:33:38.269 --> 00:33:46.069 In this first example, the largest thing that divides both 15 and 25x is five. 00:33:46.069 --> 00:33:53.990 So the GCF is five. So I pull the five out, and then I divide each of the terms by that 00:33:53.990 --> 00:33:58.769 number. And so I get three plus 5x. 00:33:58.769 --> 00:34:04.990 Pause the video for a moment and see if you can find the greatest common factor of x squared 00:34:04.990 --> 00:34:08.690 y and y squared x cubed. 00:34:08.690 --> 00:34:15.648 The biggest thing that divides both x squared y and y squared x cubed is going to be x squared 00:34:15.648 --> 00:34:18.098 times y. 00:34:18.099 --> 00:34:24.030 One way to find this is to look for the power of x that smallest in each of these terms. 00:34:24.030 --> 00:34:29.149 So that's x squared. And the power of y that smallest in each of these terms is just y 00:34:29.148 --> 00:34:31.710 to the one or y. 00:34:31.710 --> 00:34:38.599 Now if I factor out the x squared y from each of the terms, that's like dividing each term 00:34:38.599 --> 00:34:44.129 by x squared y, if I divide the first term by x squared y, I just get one. If I divide 00:34:44.128 --> 00:34:50.878 the second term by x squared y, I'm going to be left with an X and a Y. I'll write this 00:34:50.878 --> 00:34:57.199 out on the side just to make it more clear, y squared x cubed over x squared y. That's 00:34:57.199 --> 00:34:59.980 like two y's on the end. 00:34:59.980 --> 00:35:07.838 Three x's on the top and two x's and a y on the bottom. So I'm left with just an X and 00:35:07.838 --> 00:35:16.130 a Y. So I'll write the x, y here, and I factored my expression. As always, I can check my answer 00:35:16.130 --> 00:35:23.329 by multiplying out. So if I multiply out my factored expression, I get x squared y is 00:35:23.329 --> 00:35:28.849 the first term and the second term, I get x there. Now let's see three X's multiplied 00:35:28.849 --> 00:35:35.710 together and two y's multiplied together. And that checks out with what I started with. 00:35:35.710 --> 00:35:40.880 The next technique of factoring, I'd like to go over his factoring by grouping. In this 00:35:40.880 --> 00:35:46.940 example, notice that we have four terms, factoring by grouping is a handy method to look at. 00:35:46.940 --> 00:35:50.200 If you have four terms in your expression, you need to factor 00:35:50.199 --> 00:35:56.368 in order to factor by grouping, I'm first going to factor out the greatest common factor 00:35:56.369 --> 00:36:00.820 of the first two terms, and then separately, factor out the greatest common factor of the 00:36:00.820 --> 00:36:08.318 last two terms. The greatest common factor of x cubed, and 3x squared is x squared. So 00:36:08.318 --> 00:36:14.518 I factor out the x squared, and I get x plus three. And now the greatest common factor 00:36:14.518 --> 00:36:23.939 of forex and 12 is just four. So I factor out the four from those two terms. 00:36:23.940 --> 00:36:31.009 Notice that the factor of x plus three now appears in both pieces. So I can factor out 00:36:31.009 --> 00:36:36.039 the greatest common factor of x plus three, and I'll factor it out on the left side instead 00:36:36.039 --> 00:36:45.549 of the right. And now I have an x squared from this first piece, and I have a four from 00:36:45.548 --> 00:36:52.170 this second piece. And that completes my factoring by grouping. You might wonder if we could 00:36:52.170 --> 00:36:57.320 factor further by factoring the expression x squared plus four. But in fact, as we'll 00:36:57.320 --> 00:37:03.269 see later, this expression, which is a sum of two squares, x squared plus two squared 00:37:03.268 --> 00:37:06.078 does not factor any further over the integers. 00:37:06.079 --> 00:37:13.220 Next, we'll do some factoring of quadratics. a quadratic is an expression with a squared 00:37:13.219 --> 00:37:15.209 term, 00:37:15.210 --> 00:37:20.480 just a term with x in it, and a constant term with no x's in it. 00:37:20.480 --> 00:37:29.500 I'd like to factor this expression as a product of x plus or minus some number times x plus 00:37:29.500 --> 00:37:32.289 or minus some other number. 00:37:32.289 --> 00:37:37.329 The key idea is that if I can find those two numbers, then if I were to distribute out 00:37:37.329 --> 00:37:41.440 this expression, those two numbers would have to multiply to give me my constant term of 00:37:41.440 --> 00:37:48.889 eight. And these two numbers would end up having to add to give me my negative six, 00:37:48.889 --> 00:37:53.319 because when I multiply out, this number will be a coefficient of x, and this number will 00:37:53.320 --> 00:37:59.509 be also another coefficient of x, they'll add together to the negative six. 00:37:59.509 --> 00:38:04.820 So if I look at all the pairs of numbers that multiply together to give me eight, so that 00:38:04.820 --> 00:38:11.519 could be one and eight, two, and four, four and two, but that's really the same thing 00:38:11.519 --> 00:38:16.340 as I had before. And that's sort of the same thing I had before. I shouldn't forget the 00:38:16.340 --> 00:38:20.470 negatives, I could have negative one, negative eight, or I could have negative two, negative 00:38:20.469 --> 00:38:25.399 four, those alternate multiply together to give me eight. Now I just have to find, see 00:38:25.400 --> 00:38:31.579 if there's a pair of these numbers that add to negative six, and it's not hard to see 00:38:31.579 --> 00:38:39.190 that these ones will work. So now I can write out my factoring as that would be x minus 00:38:39.190 --> 00:38:45.720 two times x minus four. And it's always a good idea to check by multiplying out, I'm 00:38:45.719 --> 00:38:53.558 going to get x squared minus 4x minus 2x, plus eight. And that works out to just what 00:38:53.559 --> 00:39:00.730 I want. Now this second examples a bit more complicated, because now my leading coefficient, 00:39:00.730 --> 00:39:04.750 my coefficient of x squared is not just one, it's the number 10. 00:39:04.750 --> 00:39:08.789 Now, there are lots of different methods for approaching a problem like this. And I'm just 00:39:08.789 --> 00:39:13.889 going to show you one method, my favorite method that uses factoring by grouping, but 00:39:13.889 --> 00:39:18.139 but to start out, I'm going to multiply my coefficient of x squared by my constant term, 00:39:18.139 --> 00:39:25.548 so I'm multiplying 10 by negative six, that gives me negative 60. And I'll also take my 00:39:25.548 --> 00:39:31.690 coefficient of x the number 11. And write that down here. Now I'm going to look for 00:39:31.690 --> 00:39:38.280 two numbers that multiply to give me negative 60. And add to give me 11. You might notice 00:39:38.280 --> 00:39:42.410 that this is exactly what we were doing in the previous problem. It's just here, we didn't 00:39:42.409 --> 00:39:47.210 have to multiply the coefficient of x squared by eight, because the coefficient of x squared 00:39:47.210 --> 00:39:53.769 was just one. So to find the two numbers that multiply to negative 60 and add to 11, you 00:39:53.769 --> 00:39:58.528 might just be able to come up with them in your head thinking about it, but if not, you 00:39:58.528 --> 00:40:00.550 can figure it out. Pretty simple. 00:40:00.550 --> 00:40:05.528 thematically by writing out all the factors, pairs of factors that multiply to negative 00:40:05.528 --> 00:40:12.579 60. So I can start with negative one and 60, negative two and 30, negative three and 20. 00:40:12.579 --> 00:40:19.960 And keep going like this until I have found factors that actually add together to give 00:40:19.960 --> 00:40:28.318 me the number 11. And, and now that I look at it, I've already found them. 15 minus four, 00:40:28.318 --> 00:40:34.159 gives me 11. So I don't have to continue with my chart of factors. Now, once I found those 00:40:34.159 --> 00:40:40.858 factors, I write out my expression 10x squared, but instead of writing 11x, I write negative 00:40:40.858 --> 00:40:49.210 4x plus 15x. Now I copy down the negative six, notice that negative 4x plus 15x equals 00:40:49.210 --> 00:40:55.679 11x. That's how I chose those numbers. And so this expression is evaluates is the same 00:40:55.679 --> 00:41:01.018 as, as this expression, I haven't changed my expression. But I have turned it into something 00:41:01.018 --> 00:41:05.250 that I can apply factoring by grouping on Look, I've got four terms here. And so if 00:41:05.250 --> 00:41:11.139 I factor out my greatest common factor of my first two terms, that's let's see, I think 00:41:11.139 --> 00:41:18.009 it's 2x. So I factor out the 2x, I get 5x minus two, and then I factor out the greatest 00:41:18.010 --> 00:41:25.160 common factor of 15x and negative six, that would be three, and I get a 5x minus two, 00:41:25.159 --> 00:41:31.409 again, this is working beautifully. So I have a 5x minus two in each part. And so I put 00:41:31.409 --> 00:41:38.920 the 5x minus two on the right, and I put what's left from these terms in here. So that's 2x 00:41:38.920 --> 00:41:43.710 plus three. And I have factored my expression. 00:41:43.710 --> 00:41:50.929 There are a couple special kinds of expressions that appear frequently, that it's handy to 00:41:50.929 --> 00:41:55.149 just memorize the formula for. So the first one is the difference of squares. If you see 00:41:55.150 --> 00:42:02.460 something of the form a squared minus b squared, then you can factor that as a plus b times 00:42:02.460 --> 00:42:08.230 a minus b. And let's just check that that works. If I do a plus b times a minus b and 00:42:08.230 --> 00:42:17.358 multiply that out, I get a squared minus a b plus b A minus B squared, and those middle 00:42:17.358 --> 00:42:21.710 two terms cancel out. So it gives me back the difference of squares just like I want 00:42:21.710 --> 00:42:30.590 it. So for this first example, I if I think of x squared minus 16 as x squared minus four 00:42:30.590 --> 00:42:34.170 squared, then I can see that's a difference of squares. And I can immediately write it 00:42:34.170 --> 00:42:41.329 as x plus four times x minus four. And the second example, nine p squared minus one, 00:42:41.329 --> 00:42:50.140 that's the same thing as three p squared minus one squared. So that's three p plus one times 00:42:50.139 --> 00:42:54.139 three p minus one. 00:42:54.139 --> 00:42:57.949 Notice that if I have a sum of squares, 00:42:57.949 --> 00:43:01.169 for example, 00:43:01.170 --> 00:43:09.389 x squared plus four, which is x squared plus two squared, then that does not factor. 00:43:09.389 --> 00:43:13.239 The difference of squares formula doesn't apply. And there is no formula that applies 00:43:13.239 --> 00:43:19.989 for a sum of squares. There is, however, a formula for both a difference of cubes and 00:43:19.989 --> 00:43:27.449 a sum of cubes. The difference of cubes formula, a cubed minus b cubed is a minus b times a 00:43:27.449 --> 00:43:34.879 squared plus a b plus b squared. The formula for the sum of cubes is pretty much the same, 00:43:34.880 --> 00:43:41.318 you just switch the negative and positive sign here in here. So that gives us a plus 00:43:41.318 --> 00:43:49.018 b times a squared minus a b plus b squared. As usual, you can check these formulas by 00:43:49.018 --> 00:43:55.608 multiplying out. Let's look at one example of using these formulas. Y cubed plus 27 is 00:43:55.608 --> 00:44:03.298 actually a sum of two cubes because it's y cubed plus three cubed. So I can factor it 00:44:03.298 --> 00:44:10.969 using the sum of cubes formula by plugging in y for a and three for B. That gives me 00:44:10.969 --> 00:44:18.028 y plus three times y squared minus y times three plus three squared. And I can clean 00:44:18.028 --> 00:44:23.849 that up a little bit to read y plus three times y squared minus three y plus nine. 00:44:23.849 --> 00:44:31.190 So in this video, we went over several methods of factoring. We did factoring out the greatest 00:44:31.190 --> 00:44:40.259 common factor. We did factoring by grouping. We did factoring quadratics. And we did a 00:44:40.259 --> 00:44:47.789 difference of squares. And we did a difference and a psalm of cubes 00:44:47.789 --> 00:44:54.210 and more complicated problems, you may need to apply several these techniques in order 00:44:54.210 --> 00:44:58.470 to get through a single problem. For example, you might need to start by pulling out a greatest 00:44:58.469 --> 00:45:00.328 common factor and then 00:45:00.329 --> 00:45:03.920 Add to a factoring of quadratics, or something similar. 00:45:03.920 --> 00:45:07.700 Now go forth and factor. 00:45:07.699 --> 00:45:13.848 This video gives some additional examples of factoring. Please pause the video and decide 00:45:13.849 --> 00:45:20.009 which of these first five expressions factor and which one does not. 00:45:20.009 --> 00:45:27.469 The first expression can be factored by pulling out a common factor of x from each term. 00:45:27.469 --> 00:45:34.199 So that becomes x times x plus one. 00:45:34.199 --> 00:45:41.358 The second example can be factors as a difference of two squares, since x squared minus 25 is 00:45:41.358 --> 00:45:44.630 something squared, minus something else squared. And we know that anytime we have something 00:45:44.630 --> 00:45:55.099 like a squared minus b squared, that's a plus b times a minus b. So we can factor this as 00:45:55.099 --> 00:45:58.509 X plus five times x minus five. 00:45:58.509 --> 00:46:05.139 The third one is a sum of two squares, there's no way to factor a sum of two squares over 00:46:05.139 --> 00:46:09.690 real numbers. So this is the one that does not factor. 00:46:09.690 --> 00:46:16.608 just for completeness, let's look at the next to this next one does factor by grouping. 00:46:16.608 --> 00:46:23.259 When we factor by grouping, we pull up the biggest common factor out of the first two 00:46:23.260 --> 00:46:29.720 terms, that would be an x squared, that becomes x squared times x plus two. And then we factor 00:46:29.719 --> 00:46:36.019 as much as we can add the next two terms, that would be a three times x plus two. Notice 00:46:36.019 --> 00:46:42.420 that the x plus two factor now occurs in both of the resulting terms. So we can pull that 00:46:42.420 --> 00:46:51.528 x plus two out and get x plus two times x squared plus three, 00:46:51.528 --> 00:46:57.050 we can't factor any further because x squared plus three doesn't factor. 00:46:57.050 --> 00:47:04.810 Finally, we have a quadratic, this also factors. And I like to factor these also using a factoring 00:47:04.809 --> 00:47:12.849 by grouping trick. So first, what I do is I multiply the coefficient of x squared and 00:47:12.849 --> 00:47:19.019 the constant term, five times eight is 40. I'll write that on the top of my x. Now I 00:47:19.019 --> 00:47:24.318 take the coefficient of the x term, that's negative 14, and I write that on the bottom 00:47:24.318 --> 00:47:32.480 part of the x. Now I'm looking for two numbers that multiply to 40 and add to negative 14. 00:47:32.480 --> 00:47:37.659 Sometimes I can just guess numbers like this, but if not, I start writing out factors of 00:47:37.659 --> 00:47:46.868 40. So factors of 40, I could do one times 40. Well, now I just noticed, I'm trying to 00:47:46.869 --> 00:47:51.568 add to a negative number. So if I use two positive factors, there's no way there's going 00:47:51.568 --> 00:47:56.929 to add to a negative number. It's better for me to use negative numbers, that factor 40, 00:47:56.929 --> 00:48:02.278 a negative times a negative still multiplies to 40. But they have a chance of adding to 00:48:02.278 --> 00:48:06.460 a negative number. So but negative one and negative 40, of course, don't work, they don't 00:48:06.460 --> 00:48:11.220 add to negative 14, they add to negative 41. So let me try some other factors. The next 00:48:11.219 --> 00:48:15.838 biggest number that divides 40, besides one is two, so I'll try negative two and negative 00:48:15.838 --> 00:48:22.518 20. Those add to negative 22. That doesn't work. Next one that divides 40 would be four, 00:48:22.518 --> 00:48:29.818 so I'll try negative four and negative 10. Aha, we have a winner. So negative four plus 00:48:29.818 --> 00:48:35.460 negative 10 is negative 14, negative four times negative 10 is positive 40. We've got 00:48:35.460 --> 00:48:41.869 it. Alright, so the next step is to use factoring by grouping, we're going to first split up 00:48:41.869 --> 00:48:50.809 this negative 14x as negative 4x minus 10x. And carry down the eight and the 5x squared. 00:48:50.809 --> 00:48:57.300 Notice that this works, because I picked negative four and negative 10 to add up to negative 00:48:57.300 --> 00:49:04.680 14, so so negative 4x minus 10x, will add up to negative 14x. So I've got the same expression, 00:49:04.679 --> 00:49:09.528 just just expand it out a little bit. Now I have four terms, I can do factoring by grouping, 00:49:09.528 --> 00:49:13.989 so I can group the first two terms and factor out the biggest thing I can that will be n 00:49:13.989 --> 00:49:19.348 x times 5x minus four. And now I'll factor the biggest thing I can out of these two numbers, 00:49:19.349 --> 00:49:26.499 including the the negative. So that becomes, let's see, I can factor out a negative two 00:49:26.498 --> 00:49:33.588 and that becomes 5x minus four since negative two times minus four is eight. All right, 00:49:33.588 --> 00:49:40.068 I've got the same 5x plus four in both my terms, so factoring by grouping is going swimmingly, 00:49:40.068 --> 00:49:45.858 I can factor out the 5x minus four from both those terms and I get the x minus two, and 00:49:45.858 --> 00:49:51.440 I factored this quadratic. If I want to, of course, I can always check my work by distributing 00:49:51.440 --> 00:49:59.179 out by multiplying out. So a check here would be multiplying 5x times x is 5x. 00:49:59.179 --> 00:50:08.288 squared 5x minus 10 to minus two is minus 10x minus four times x is minus 4x. And minus 00:50:08.289 --> 00:50:13.460 four times minus two is plus eight. So let's see, this does check out to exactly what it 00:50:13.460 --> 00:50:22.119 should be. So that was the method of factoring a quadratic. 00:50:22.119 --> 00:50:27.358 And all of these factors except for the sum of squares. 00:50:27.358 --> 00:50:35.009 So we saw that factoring by grouping is handy for factoring this expression here. It was 00:50:35.009 --> 00:50:39.929 also handy for factoring the quadratic indirectly, after splitting up 00:50:39.929 --> 00:50:44.389 the middle term 00:50:44.389 --> 00:50:52.279 into two terms. So how can you tell when a an expression is is appropriate to factor 00:50:52.280 --> 00:50:57.690 by grouping, there's, there's an easy way to tell that it might be a candidate, and 00:50:57.690 --> 00:51:00.099 that's that it has four terms. 00:51:00.099 --> 00:51:05.539 So if you see four terms, or in the case of quadratic, you can split it up into four terms, 00:51:05.539 --> 00:51:09.030 then that's a good candidate for factoring by grouping because you can group the first 00:51:09.030 --> 00:51:16.210 two terms group the second two terms, factoring by grouping all always work on, on expressions 00:51:16.210 --> 00:51:21.559 with four terms. But, but that's like the first thing to look for. So let's just review 00:51:21.559 --> 00:51:25.829 what are the same main techniques of factoring. We saw these on the previous page, we saw 00:51:25.829 --> 00:51:32.230 there was pull out common factors. There's difference of squares. 00:51:32.230 --> 00:51:36.230 There's factoring by grouping. 00:51:36.230 --> 00:51:41.849 There's factoring quadratics. 00:51:41.849 --> 00:51:49.970 And one more that I didn't mention is factoring sums and differences of cubes. 00:51:49.969 --> 00:51:57.328 that uses the formulas, aq minus b cubed is a minus b times a squared plus a b plus b 00:51:57.329 --> 00:52:07.930 squared. And a cubed plus b cubed is a plus b times a squared minus a b plus b squared. 00:52:07.929 --> 00:52:11.068 One important tip when factoring, 00:52:11.068 --> 00:52:17.150 I always recommend doing this first, pull out the common factors first. 00:52:17.150 --> 00:52:21.798 That'll simplify things and making the rest of factoring easier. One more tip is that 00:52:21.798 --> 00:52:28.130 you might need to do several these factoring techniques in one problem, for example, 00:52:28.130 --> 00:52:31.869 you might have to first pull out a common factor, then factor a difference of squares. 00:52:31.869 --> 00:52:35.338 And then you might notice that one of your factors is itself a difference of squares, 00:52:35.338 --> 00:52:41.038 and you have to apply a difference of squares again, so don't stop when you factor a little 00:52:41.039 --> 00:52:43.910 bit, keep factoring as far as you can go. 00:52:43.909 --> 00:52:49.411 Here are some extra examples of factoring quadratics. For you to practice, please pause 00:52:49.411 --> 00:52:52.619 the video and give these a try. 00:52:52.619 --> 00:53:01.130 For the first one, let's multiply two times negative 14. That gives us negative 28. And 00:53:01.130 --> 00:53:06.338 then we'll bring the three down in the bottom of the x. Now we're looking for two numbers 00:53:06.338 --> 00:53:14.269 that multiply to negative 28 and add to three. Well, to multiply two numbers to get a negative 00:53:14.269 --> 00:53:23.518 28, we'll need one of them to be negative and one of them to be positive. So to be one, 00:53:23.518 --> 00:53:26.808 negative 128, or one, negative 28, those don't work. Let's see negative 214 or two, negative 00:53:26.809 --> 00:53:30.309 14, those don't work. Hey, I just noticed the positive number had better be bigger than 00:53:30.309 --> 00:53:37.400 the negative number, so they add to a positive number. Let's see what comes next. How about 00:53:37.400 --> 00:53:43.460 negative four times seven, four times negative seven, I think four, negative four times seven 00:53:43.460 --> 00:53:49.199 will work. So I'll write those here at negative four, seven, 00:53:49.199 --> 00:53:58.009 copy down the two z squared. And I'll split up the three z into negative four z plus seven 00:53:58.009 --> 00:54:05.170 z and then minus 14. Now factoring by grouping, pull out a to z, that becomes z minus two, 00:54:05.170 --> 00:54:12.650 pull out a seven and that becomes z minus two again, looking good. I've got two z plus 00:54:12.650 --> 00:54:18.259 seven times z minus two as my factored expression. 00:54:18.259 --> 00:54:23.048 My second expression, I could work at the same way, drawing my axe and factoring by 00:54:23.048 --> 00:54:28.619 grouping that kind of thing. But it's actually going to be easier if I notice first that 00:54:28.619 --> 00:54:32.170 I can pull out a common factor from all of my terms, though, that'll make things a lot 00:54:32.170 --> 00:54:36.130 simpler to deal with. So notice that a five divides each of these terms, and in fact, 00:54:36.130 --> 00:54:39.660 I'm going to go ahead and pull out the negative five because I don't like having negatives 00:54:39.659 --> 00:54:44.690 in front of my squared term. So I'm going to pull out a common factor of negative five. 00:54:44.690 --> 00:54:49.470 Again, it would work if I forgot to do this, but it would be a lot more complicated. So 00:54:49.469 --> 00:54:56.699 negative five v squared, this becomes minus, this becomes plus nine V, since nine times 00:54:56.699 --> 00:54:59.868 negative five is negative 45. And this becomes minus 00:54:59.869 --> 00:55:06.650 10 cents native 10 times negative five is positive 50. Now I can start my x and my factoring 00:55:06.650 --> 00:55:10.849 by grouping, or I can use kind of a shortcut method, which you may have seen before. So 00:55:10.849 --> 00:55:15.588 I can just put these here, and then I know that whatever numbers go here, they're gonna 00:55:15.588 --> 00:55:21.150 have to multiply to the negative 10. And they're gonna have to add to the nine. So that would 00:55:21.150 --> 00:55:26.289 be plus 10, and a minus one will do the trick. 00:55:26.289 --> 00:55:32.230 Those are all my factoring examples for today. I hope you enjoy your snow morning and have 00:55:32.230 --> 00:55:37.929 a chance to spend some time working in ALEKS. Bye. 00:55:37.929 --> 00:55:43.028 This video is about working with rational expressions. A rational expression is a fraction 00:55:43.028 --> 00:55:49.460 usually with variables in it, something like x plus two over x squared minus three is a 00:55:49.460 --> 00:55:55.360 rational expression. In this video, we'll practice adding, subtracting, multiplying 00:55:55.360 --> 00:56:01.660 and dividing rational expressions and simplifying them to lowest terms. 00:56:01.659 --> 00:56:07.808 We'll start with simplifying to lowest terms. Recall that if you have a fraction with just 00:56:07.809 --> 00:56:14.599 numbers in it, something like 21 over 45, we can reduce it to lowest terms by factoring 00:56:14.599 --> 00:56:17.430 the numerator 00:56:17.429 --> 00:56:22.199 and factoring the denominator 00:56:22.199 --> 00:56:26.019 and then canceling common factors. 00:56:26.019 --> 00:56:35.478 So in this example, the three is cancel, and our fraction reduces to seven over 15. 00:56:35.478 --> 00:56:40.259 If we want to reduce a rational expression with the variables and add to lowest terms, 00:56:40.259 --> 00:56:47.309 we proceed the same way. First, we'll factor the numerator, that's three times x plus two, 00:56:47.309 --> 00:56:54.680 and then factor the denominator. In this case of factors 2x plus two times x plus two, we 00:56:54.679 --> 00:57:00.710 could also write that as x plus two squared. Now we cancel the common factors. And we're 00:57:00.710 --> 00:57:08.579 left with three over x plus two. Definitely a simpler way of writing that rational expression. 00:57:08.579 --> 00:57:15.900 Next, let's practice multiplying and dividing. Recall that if we multiply two fractions with 00:57:15.900 --> 00:57:20.789 just numbers in them, we simply multiply the numerators and multiply the denominators. 00:57:20.789 --> 00:57:28.599 So in this case, we would get four times two over three times five or 8/15. 00:57:28.599 --> 00:57:35.338 If we want to divide two fractions, like in the second example, then we can rewrite it 00:57:35.338 --> 00:57:43.170 as multiplying by the reciprocal of the fraction on the denominator. So here, we get four fifths 00:57:43.170 --> 00:57:50.970 times three halves, and that gives us 12 tenths. But actually, we could reduce that fraction 00:57:50.969 --> 00:57:53.818 to six fifths, 00:57:53.818 --> 00:58:00.000 we use the same rules when we compute the product or quotient of two rational expressions 00:58:00.000 --> 00:58:06.548 with the variables. And then here, we're trying to divide two rational expressions. So instead, 00:58:06.548 --> 00:58:12.889 we can multiply by the reciprocal. I call this flipping and multiplying. 00:58:12.889 --> 00:58:18.828 And now we just multiply the numerators. 00:58:18.829 --> 00:58:21.859 And multiply the denominators. 00:58:21.858 --> 00:58:28.889 It might be tempting at this point to multiply out to distribute out the numerator and the 00:58:28.889 --> 00:58:32.748 denominator. But actually, it's better to leave it in this factored form and factored 00:58:32.748 --> 00:58:38.139 even more completely. That way, we'll be able to reduce the rational expression to cancel 00:58:38.139 --> 00:58:46.048 the common factors. So let's factor even more the x squared plus x factors as x times x 00:58:46.048 --> 00:58:52.440 plus one, and x squared minus 16. And that's a difference of two squares, that's x plus 00:58:52.440 --> 00:58:59.510 four times x minus four, the denominator is already fully factored, so we'll just copy 00:58:59.510 --> 00:59:06.569 it over. And now we can cancel common factors here and here, and we're left with x times 00:59:06.568 --> 00:59:12.659 x minus four. This is our final answer. 00:59:12.659 --> 00:59:16.170 Adding and subtracting fractions is a little more complicated because we first have to 00:59:16.170 --> 00:59:23.088 find a common denominator. A common denominator is an expression that both denominators divided 00:59:23.088 --> 00:59:29.190 into, it's usually best of the long run to use the least common denominator, which is 00:59:29.190 --> 00:59:34.028 the smallest expression that both denominators divided into. 00:59:34.028 --> 00:59:39.219 In this example, if we just want a common denominator, we could use six times 15, which 00:59:39.219 --> 00:59:46.058 is 90 because both six and 15 divided evenly into 90. But if we want the least common denominator, 00:59:46.059 --> 00:59:55.150 the best way to do that is to factor the two denominators. So six is two times 315 is three 00:59:55.150 --> 00:59:59.900 times five, and then put together only the factors we need for 00:59:59.900 --> 01:00:07.789 Both six and 50 into divider numbers. So if we just use two times three times five, which 01:00:07.789 --> 01:00:14.159 is 30, we know that two times three will divide it, and three times five will also divide 01:00:14.159 --> 01:00:19.440 it. And we won't be able to get a denominator any smaller, because we need the factors two, 01:00:19.440 --> 01:00:24.970 three, and five, in order to ensure both these numbers divided. Once we have our least common 01:00:24.969 --> 01:00:31.598 denominator, we can rewrite each of our fractions in terms of that denominator. So seven, six, 01:00:31.599 --> 01:00:38.940 I need to get a 30 in the denominator, so I'm going to multiply that by five over five, 01:00:38.940 --> 01:00:45.539 and multiply by the factors that are missing from the current denominator in order to get 01:00:45.539 --> 01:00:54.068 my least common denominator of 30. For the second fraction, for 15th 15 times two is 01:00:54.068 --> 01:00:57.829 30. So I'm going to multiply by two over two, 01:00:57.829 --> 01:01:08.971 I can rewrite this as 3530 s minus 8/30. And now that I have a common denominator, I can 01:01:08.971 --> 01:01:15.608 just subtract my two numerators. And I get 27/30. 01:01:15.608 --> 01:01:20.119 If I factor, I can reduce this 01:01:20.119 --> 01:01:28.108 to three squared over two times five, which is nine tenths. The process for finding the 01:01:28.108 --> 01:01:33.759 sum of two rational expressions with variables in them follows the exact same process. First, 01:01:33.759 --> 01:01:39.440 we have to find the least common denominator, I'll do that by factoring the two denominators. 01:01:39.440 --> 01:01:45.659 So 2x plus two factors as two times x plus 1x squared minus one, that's a difference 01:01:45.659 --> 01:01:51.858 of two squares. So that's x plus one times x minus one. Now for the least common denominator, 01:01:51.858 --> 01:01:57.558 I'm going to take all the factors, I need to get an expression that each of these divides 01:01:57.559 --> 01:02:02.329 into, so I need the factor two, I need the factor x plus one, and I need the factor x 01:02:02.329 --> 01:02:08.010 minus one, I don't have to repeat the factor x plus one I just need to have at one time. 01:02:08.010 --> 01:02:14.410 And so I will get my least common denominator two times x plus one times x minus one, I'm 01:02:14.409 --> 01:02:17.868 not going to bother multiplying this out, it's actually better to leave it in factored 01:02:17.869 --> 01:02:24.920 form to help me simplify later. Now I can rewrite each of my two rational expressions 01:02:24.920 --> 01:02:29.950 by multiplying by whatever's missing from the denominator in terms of the least common 01:02:29.949 --> 01:02:36.098 denominator. So what I mean is, I can rewrite three over 2x plus two, I'll write the 2x 01:02:36.099 --> 01:02:41.278 plus two is two times x plus one, I'll write it in factored form. And then I noticed that 01:02:41.278 --> 01:02:45.778 compared to the least common denominator, I'm missing the factor of x minus one. So 01:02:45.778 --> 01:02:52.099 I multiply the numerator and the denominator by x minus one, I just need it in the denominator. 01:02:52.099 --> 01:02:56.660 But I can't get away with just multiplying by the denominator without changing my expression, 01:02:56.659 --> 01:03:00.788 I have to multiply by it on the numerator and the denominator. So I'm just multiplying 01:03:00.789 --> 01:03:06.690 by one and a fancy form and not changing the value here. So now I do the same thing for 01:03:06.690 --> 01:03:10.950 the second rational expression, I'll I'll write the denominator in factored form to 01:03:10.949 --> 01:03:16.989 make it easier to see what's missing from the denominator. What's missing in this denominator, 01:03:16.989 --> 01:03:21.588 compared to my least common denominator is just the factor two, so multiply the numerator 01:03:21.588 --> 01:03:27.590 and the denominator by two. Now I can rewrite everything. So the first rational expression 01:03:27.590 --> 01:03:36.059 becomes three times x minus one over two times x plus 1x minus one, and the second one becomes 01:03:36.059 --> 01:03:44.849 five times two over two times x plus 1x minus one, notice that I now have a common denominator. 01:03:44.849 --> 01:03:51.519 So I can just add together my numerators. So I get three times x minus one plus 10 over 01:03:51.518 --> 01:03:58.228 two times x plus 1x minus one. I'd like to simplify this. And the best way to do that 01:03:58.228 --> 01:04:04.218 is to leave the denominator in factored form. But I do have to multiply out the numerator 01:04:04.219 --> 01:04:12.599 so that I can add things together. So I get 3x minus three plus 10 over two times x plus 01:04:12.599 --> 01:04:22.130 1x minus one, or 3x plus seven over two times x plus 1x minus one. Now 3x plus seven doesn't 01:04:22.130 --> 01:04:29.160 factor. And there's therefore no factors that I can cancel out. So this is already reduced. 01:04:29.159 --> 01:04:33.129 As much as it can be. This is my final answer. 01:04:33.130 --> 01:04:40.778 In this video, we saw how to simplify rational expressions to lowest terms by factoring and 01:04:40.778 --> 01:04:42.980 canceling common factors. 01:04:42.980 --> 01:04:47.949 We also saw how to multiply rational expressions by multiplying the numerator and multiplying 01:04:47.949 --> 01:04:53.538 the denominator, how to divide rational expressions by flipping and multiplying and how to add 01:04:53.539 --> 01:05:00.579 and subtract rational expressions by writing them in terms of the least common denominator 01:05:00.579 --> 01:05:07.119 This video is about solving quadratic equations. a quadratic equation is an equation that contains 01:05:07.119 --> 01:05:11.960 the square of the variable, say x squared, but no higher powers of x. 01:05:11.960 --> 01:05:20.449 The standard form for a quadratic equation is the form a x squared plus b x plus c equals 01:05:20.449 --> 01:05:25.498 zero, where A, B and C 01:05:25.498 --> 01:05:31.419 represent real numbers. And a is not zero so that we actually have an x squared term. 01:05:31.420 --> 01:05:40.450 Let me give you an example. 3x squared plus 7x minus two equals zero is a quadratic equation 01:05:40.449 --> 01:05:49.788 in standard form, here a is three, B is seven, and C is minus two. The equation 3x squared 01:05:49.789 --> 01:05:56.839 equals minus 7x plus two is also a quadratic equation, it's just not in standard form. 01:05:56.838 --> 01:06:02.449 The key steps to solving quadratic equations are usually to write the equation in standard 01:06:02.449 --> 01:06:07.449 form, and then either factor it 01:06:07.449 --> 01:06:14.548 or use the quadratic formula, which I'll show you later in this video. 01:06:14.548 --> 01:06:21.119 Let's start with the example y squared equals 18 minus seven y. Here our variable is y, 01:06:21.119 --> 01:06:26.650 and we need to rewrite this quadratic equation in standard form, we can do this by subtracting 01:06:26.650 --> 01:06:33.680 18 from both sides and adding seven y to both sides. That gives us the equation y squared 01:06:33.679 --> 01:06:40.848 minus 18 plus seven y equals zero. And I can rearrange a little bit to get y squared plus 01:06:40.849 --> 01:06:48.300 seven y minus 18 equals zero. Now I've got my equation in standard form. Next, I'm going 01:06:48.300 --> 01:06:54.589 to try to factor it. So I need to look for two numbers that multiply to negative 18 and 01:06:54.588 --> 01:06:57.489 add to seven. 01:06:57.489 --> 01:07:02.659 two numbers that work are nine and negative two, so I can factor my expression on the 01:07:02.659 --> 01:07:12.039 left as y plus nine times y minus two equals zero. Now, anytime you have two quantities 01:07:12.039 --> 01:07:17.389 that multiply together to give you zero, either the first quantity has to be zero, or the 01:07:17.389 --> 01:07:22.478 second quantity has to be zero, or I suppose they both could be zero. In some situations, 01:07:22.478 --> 01:07:27.879 this is really handy, because that means that I know that either y plus nine equals zero, 01:07:27.880 --> 01:07:36.410 or y minus two equals zero. So I can as my next step, set my factors equal to zero. So 01:07:36.409 --> 01:07:43.278 y plus nine equals zero, or y minus two equals zero, which means that y equals negative nine, 01:07:43.278 --> 01:07:46.619 or y equals two. 01:07:46.619 --> 01:07:51.039 It's not a bad idea to check that those answers actually work by plugging them into the original 01:07:51.039 --> 01:07:56.998 equation, negative nine squared, does that equal 18 minus seven times negative nine, 01:07:56.998 --> 01:08:04.459 and you can work out that it does. And similarly, two squared equals 18 minus seven times two. 01:08:04.460 --> 01:08:11.449 In the next example, let's find solutions of the equation w squared equals 121. This 01:08:11.449 --> 01:08:17.729 is a quadratic equation, because it's got a square of my variable W, I can rewrite it 01:08:17.729 --> 01:08:24.278 in standard form by subtracting 121 from both sides. 01:08:24.279 --> 01:08:31.839 Notice that A is equal to one b is equal to zero because there's no w term, and c is equal 01:08:31.838 --> 01:08:39.340 to negative 121 in the standard form, a W squared plus BW plus c equals zero. Next step, 01:08:39.340 --> 01:08:47.980 I'm going to try to factor this expression. So since 121, is 11 squared, this is a difference 01:08:47.979 --> 01:08:56.129 of two squares, and it factors as w plus 11, times w minus 11 is equal to zero. If I set 01:08:56.130 --> 01:08:59.560 the factors equal to zero, 01:08:59.560 --> 01:09:07.560 I get w plus 11 equals zero or w minus 11 equals zero. So w equals minus 11, or w equals 01:09:07.560 --> 01:09:14.289 11. In this example, I could have solved the equation more simply, I could have instead 01:09:14.289 --> 01:09:20.539 said that if W squared is 121, and then w has to equal plus or minus the square root 01:09:20.539 --> 01:09:26.189 of 121. In other words, W is plus or minus 11. 01:09:26.189 --> 01:09:30.939 If you saw the equation this way, it's important to remember the plus or minus since minus 01:09:30.939 --> 01:09:36.019 11 squared equals 121, just like 11 squared does. 01:09:36.020 --> 01:09:41.790 Now let's find the solutions for the equation. x times x plus two equals seven. Some people 01:09:41.789 --> 01:09:46.630 might be tempted to say that, oh, if two numbers multiply to equal seven, then one of them 01:09:46.630 --> 01:09:51.529 better equal one and the other equals seven or maybe negative one and negative seven. 01:09:51.529 --> 01:09:57.590 But that's faulty reasoning in this case, because x and x plus two don't have to be 01:09:57.590 --> 01:10:00.489 whole numbers. They could be crazy. 01:10:00.489 --> 01:10:08.729 fractions or even irrational numbers. So instead, let's rewrite this equation in standard form. 01:10:08.729 --> 01:10:16.879 To do that, I'm first going to multiply out. So x times x is x squared x times two is 2x. 01:10:16.880 --> 01:10:23.310 That equals seven, and I'll subtract the seven from both sides to get x squared plus 2x minus 01:10:23.310 --> 01:10:28.770 seven is zero. Now I'm looking to factor it. So I need two numbers that multiply to negative 01:10:28.770 --> 01:10:34.770 seven and add to two, since the only way to factor negative seven is as negative one times 01:10:34.770 --> 01:10:39.580 seven or seven times negative one, it's easy to see that there are no whole numbers that 01:10:39.579 --> 01:10:46.750 will do will work. So there's no way to factor this expression over the integers. Instead, 01:10:46.750 --> 01:10:53.140 let's use the quadratic equation. So we have our leading coefficient of x squared is one. 01:10:53.140 --> 01:11:00.000 So A is one, B is two, and C is minus seven. And we're going to plug that into the equation 01:11:00.000 --> 01:11:06.869 quadratic equation, which goes x equals negative b plus or minus the square root of b squared 01:11:06.869 --> 01:11:10.279 minus four, I see all over two 01:11:10.279 --> 01:11:15.059 different people have different ways of remembering this formula, I'd like to remember it by seeing 01:11:15.060 --> 01:11:22.420 it x equals negative b plus or minus the square root of b squared minus four, I see oh over 01:11:22.420 --> 01:11:28.060 to a, but you can use any pneumonic you like. Anyway, plugging in here, we have x equals 01:11:28.060 --> 01:11:34.140 negative two plus or minus the square root of two squared minus four times one times 01:11:34.140 --> 01:11:40.730 negative seven support and remember the negative seven there to all over two times one. 01:11:40.729 --> 01:11:48.429 Now two squared is four, and four times one times negative seven is negative 28. So this 01:11:48.430 --> 01:11:55.360 whole quantity under the square root sign becomes four minus negative 28, or 32. So 01:11:55.359 --> 01:12:00.849 I can rewrite this as x equals negative two plus or minus the square root of 32, all over 01:12:00.850 --> 01:12:01.890 two. 01:12:01.890 --> 01:12:11.079 Since 32, is 16 times two and 16 is a perfect square, I can rewrite this as negative two 01:12:11.079 --> 01:12:15.819 plus or minus the square root of 16 times the square root of two over two, which is 01:12:15.819 --> 01:12:19.920 negative two plus or minus four times the square root of two over two. 01:12:19.920 --> 01:12:26.159 Next, I'm going to split out my fraction 01:12:26.159 --> 01:12:32.159 as negative two over two plus or minus four square root of two over two, and then simplify 01:12:32.159 --> 01:12:38.729 those fractions. This becomes negative one plus or minus two square root of two. So my 01:12:38.729 --> 01:12:45.039 answers are negative one plus two square root of two and negative one minus two square root 01:12:45.039 --> 01:12:52.039 of two. And if I need a decimal answer for any reason, I could work this out on my calculator. 01:12:52.039 --> 01:12:57.390 As our final example, let's find all real solutions for the equation, one half y squared 01:12:57.390 --> 01:13:04.600 equals 1/3 y minus two. I'll start as usual by putting it in standard form. So that gives 01:13:04.600 --> 01:13:12.810 me one half y squared minus 1/3 y plus two equals zero, I could go ahead and start trying 01:13:12.810 --> 01:13:17.940 to factor or use the quadratic formula right now. But I find fractional coefficients kind 01:13:17.939 --> 01:13:22.649 of annoying. So I'd like to get rid of them. By doing what I call clearing the denominator, 01:13:22.649 --> 01:13:27.879 that means I'm going to multiply the whole entire equation by the least common denominator. 01:13:27.880 --> 01:13:33.020 In this case, the least common denominator is two times three or six. So I'll multiply 01:13:33.020 --> 01:13:37.300 the whole equation by six have to make sure I multiplied both sides of the equation, but 01:13:37.300 --> 01:13:42.220 in this case, six times zero is just zero. And when I distribute the six, I get three 01:13:42.220 --> 01:13:47.140 y squared minus two y plus 12 equals zero. 01:13:47.140 --> 01:13:51.760 Now, I could try to factor this, but I think it's easier probably just to plunge in and 01:13:51.760 --> 01:13:54.000 use the quadratic formula. 01:13:54.000 --> 01:14:01.909 So I get x equals negative B, that's negative negative two or two, plus or minus the square 01:14:01.909 --> 01:14:12.539 root of b squared, minus four times a times c, all over to a. 01:14:12.539 --> 01:14:17.590 Working out the stuff in a square root sign, negative two squared is four. And here we 01:14:17.590 --> 01:14:21.699 have, let's see 144. 01:14:21.699 --> 01:14:28.449 So this simplifies to x equals to plus or minus the square root of four minus 144. That's 01:14:28.449 --> 01:14:35.210 negative 140. All of our sex. Well, if you're concerned about that negative number under 01:14:35.210 --> 01:14:40.399 the square root sign, you should be we can't take the square root of a negative number 01:14:40.399 --> 01:14:45.469 and get an N get a real number is our answer. There's no real number whose square is a negative 01:14:45.470 --> 01:14:54.650 number. And therefore, our conclusion is we have no real solutions to this quadratic equation. 01:14:54.649 --> 01:14:59.689 In this video, we solve some quadratic equations by first writing them in standard form and 01:14:59.689 --> 01:15:00.689 then 01:15:00.689 --> 01:15:03.669 Either factoring or using the quadratic formula. 01:15:03.670 --> 01:15:08.440 In some examples, factoring doesn't work, it's not possible to factor the equation. 01:15:08.439 --> 01:15:13.969 But in fact, using the quadratic formula will always work even if it's also possible to 01:15:13.970 --> 01:15:19.011 solve it by factoring. So you can't really lose by using the quadratic formula. It's 01:15:19.011 --> 01:15:23.000 just sometimes it'll be faster to factor instead. 01:15:23.000 --> 01:15:29.539 This video is about solving rational equations. A rational equation, like this one is an equation 01:15:29.539 --> 01:15:34.189 that has rational expressions in that, in other words, an equation that has some variables 01:15:34.189 --> 01:15:35.359 in the denominator. 01:15:35.359 --> 01:15:41.069 There are several different approaches for solving a rational equation, but they all 01:15:41.069 --> 01:15:46.719 start by finding the least common denominator. In this example, the denominators are x plus 01:15:46.720 --> 01:15:52.350 three and x, we can think of one as just having a denominator of one. 01:15:52.350 --> 01:15:57.250 Since the denominators don't have any factors in common, I can find the least common denominator 01:15:57.250 --> 01:16:00.689 just by multiplying them together. 01:16:00.689 --> 01:16:04.359 My next step is going to be clearing the denominator. 01:16:04.359 --> 01:16:12.460 By this, I mean that I multiply both sides of my equation by this least common denominator, 01:16:12.460 --> 01:16:19.270 x plus three times x, I multiply on the left side of the equation, and I multiply by the 01:16:19.270 --> 01:16:22.310 same thing on the right side of the equation. 01:16:22.310 --> 01:16:28.250 Since I'm doing the same thing to both sides of the equation, I don't change the the value 01:16:28.250 --> 01:16:33.020 of the equation. Multiplying the least common denominator on both sides of the equation 01:16:33.020 --> 01:16:38.650 is equivalent to multiplying it by all three terms in the equation, I can see this when 01:16:38.649 --> 01:16:41.189 I multiply out, 01:16:41.189 --> 01:16:47.909 I'll rewrite the left side the same as before, pretty much. And then I'll distribute the 01:16:47.909 --> 01:16:56.029 right side to get x plus three times x times one plus x plus three times x times one over 01:16:56.029 --> 01:17:01.920 x. So I've actually multiplied the least common denominator by all three terms of my equation. 01:17:01.920 --> 01:17:07.920 Now I can have a blast canceling things. The x plus three cancels with the x plus three 01:17:07.920 --> 01:17:10.119 on the denominator. 01:17:10.119 --> 01:17:15.579 The here are nothing cancels out because there's no denominator, and here are the x in the 01:17:15.579 --> 01:17:18.470 numerator cancels with the x in the denominator. 01:17:18.470 --> 01:17:26.310 So I can rewrite my expression as x squared equals x plus three times x times one plus 01:17:26.310 --> 01:17:29.550 x plus three. Now I'm going to simplify. 01:17:29.550 --> 01:17:36.690 So I'll leave the x squared alone on this side, I'll distribute out x squared plus 3x 01:17:36.689 --> 01:17:44.039 plus x plus three, hey, look, the x squared is cancel on both sides. And so I get zero 01:17:44.039 --> 01:17:51.199 equals 4x plus three, so 4x is negative three, and x is negative three fourths. Finally, 01:17:51.199 --> 01:17:57.409 I'm going to plug in my answer to check. This is a good idea for any kind of equation. But 01:17:57.409 --> 01:18:01.609 it's especially important for a rational equation because occasionally for rational equations, 01:18:01.609 --> 01:18:05.699 you'll get what's called extraneous solution solutions that don't actually work in your 01:18:05.699 --> 01:18:09.949 original equation because they make the denominator zero. Now, in this example, I don't think 01:18:09.949 --> 01:18:16.029 we're going to get any extraneous equations because negative three fourths is not going 01:18:16.029 --> 01:18:20.670 to make any of these denominators zero, so it should work out fine when I plug in. If 01:18:20.670 --> 01:18:22.590 I plug in, 01:18:22.590 --> 01:18:29.489 I get this, I can simplify 01:18:29.489 --> 01:18:34.119 the denominator here, negative three fourths plus three, three is 12 fourths, as becomes 01:18:34.119 --> 01:18:42.680 nine fourths. And this is one I'll flip and multiply to get minus four thirds. So here, 01:18:42.680 --> 01:18:49.570 I can simplify my complex fraction, it ends up being negative three nights, and one minus 01:18:49.569 --> 01:18:55.960 four thirds is negative 1/3. So that all seems to check out. 01:18:55.960 --> 01:19:00.890 And so my final answer is x equals negative three fourths. 01:19:00.890 --> 01:19:05.310 This next example looks a little trickier. And it is, but the same approach will work. 01:19:05.310 --> 01:19:13.410 First off, find the least common denominator. So here, my denominators are c minus five, 01:19:13.409 --> 01:19:22.029 c plus one, and C squared minus four c minus five, I'm going to factor that as C minus 01:19:22.029 --> 01:19:29.189 five times c plus one. Now, my least common denominator needs to have just enough factors 01:19:29.189 --> 01:19:35.269 to that each of these denominators divided into it. So I need the factor c minus five, 01:19:35.270 --> 01:19:39.960 I need the factor c plus one. And now I've already got all the factors I need for this 01:19:39.960 --> 01:19:47.480 denominator. So here is my least common denominator. Next step is to clear the denominators. 01:19:47.479 --> 01:19:53.589 So I do this by multiplying both sides of the equation by my least common denominator. 01:19:53.590 --> 01:20:01.560 In fact, I can just multiply each of the three terms by this least common denominator 01:20:01.560 --> 01:20:06.789 I went ahead and wrote my third denominator in factored form to make it easier to see 01:20:06.789 --> 01:20:16.619 what cancels. Now canceling time dies, this dies. And both of those factors die. cancelling 01:20:16.619 --> 01:20:21.229 out the denominator is the whole point of multiplying by the least common denominator, 01:20:21.229 --> 01:20:25.189 you're multiplying by something that's big enough to kill every single denominator, so 01:20:25.189 --> 01:20:28.649 you don't have to deal with denominators anymore. 01:20:28.649 --> 01:20:32.069 Now I'm going to simplify by multiplying out. 01:20:32.069 --> 01:20:41.569 So I get, let's see, c plus one times four c, that's four c squared plus four c, now 01:20:41.569 --> 01:20:51.130 I get minus just c minus five, and then over here, I get three c squared plus three, 01:20:51.130 --> 01:21:00.440 I can rewrite the minus quantity c minus five is minus c plus five. 01:21:00.439 --> 01:21:06.829 And now I can subtract the three c squared from both sides to get just a C squared over 01:21:06.829 --> 01:21:15.710 here, and the four c minus c that becomes a three C. 01:21:15.710 --> 01:21:21.829 And finally, I can subtract the three from both sides to get c squared plus three c plus 01:21:21.829 --> 01:21:28.979 two equals zero. got myself a quadratic equation that looks like a nice one that factors. So 01:21:28.979 --> 01:21:36.229 this factors to C plus one times c plus two equals zero. So either c plus one is zero, 01:21:36.229 --> 01:21:44.279 or C plus two is zero. So C equals negative one, or C equals negative two. 01:21:44.279 --> 01:21:49.689 Now let's see, we need to still check our answers. 01:21:49.689 --> 01:21:54.519 Without even going to the trouble of calculating anything, I can see that C equals negative 01:21:54.520 --> 01:22:01.160 one is not going to work, because if I plug it in to this denominator here, I get a denominator 01:22:01.159 --> 01:22:09.250 zero, which doesn't make sense. So C equals minus one is an extraneous solution, it doesn't 01:22:09.250 --> 01:22:16.789 actually satisfy my original equation. And so I can just cross it right out, C equals 01:22:16.789 --> 01:22:19.869 negative two. 01:22:19.869 --> 01:22:23.809 I can go if I go ahead, and that doesn't make any of my denominators zero. So if I haven't 01:22:23.810 --> 01:22:30.780 made any mistakes, it should satisfy my original equation, but, but I'll just plug it in to 01:22:30.779 --> 01:22:33.619 be sure. 01:22:33.619 --> 01:22:36.899 And after some simplifying, 01:22:36.899 --> 01:22:39.849 I get a true statement. 01:22:39.850 --> 01:22:46.690 So my final answer is C equals negative two. In this video, we saw the couple of rational 01:22:46.689 --> 01:22:53.349 equations using the method of finding the least common denominator and then clearing 01:22:53.350 --> 01:22:55.600 the denominator, 01:22:55.600 --> 01:22:59.490 we cleared the denominator by multiplying both sides of the equation by the least common 01:22:59.489 --> 01:23:04.389 denominator or equivalently. multiplying each of the terms by that denominator. 01:23:04.390 --> 01:23:10.250 There's another equivalent method that some people prefer, it still starts out the same, 01:23:10.250 --> 01:23:16.390 we find the least common denominator, but then we write all the fractions 01:23:16.390 --> 01:23:22.100 over that least common denominator. So in this example, we'd still use the least common 01:23:22.100 --> 01:23:27.829 denominator of x plus three times x. But our next step would be to write each of these 01:23:27.829 --> 01:23:32.350 rational expressions over that common denominator by multiplying the top and the bottom by the 01:23:32.350 --> 01:23:38.880 appropriate things. So one, in order to get the common denominator of x plus 3x, I need 01:23:38.880 --> 01:23:44.310 to multiply the top and the bottom by x plus three times x, one over x, I need to multiply 01:23:44.310 --> 01:23:49.640 the top and the bottom just by x plus three since that's what's missing from the denominator 01:23:49.640 --> 01:23:53.380 x. Now, if I simplify a little bit, 01:23:53.380 --> 01:24:00.090 let's say this is x squared over that common denominator, and 01:24:00.090 --> 01:24:09.150 here I have just x plus three times x over that denominator, and here I have x plus three 01:24:09.149 --> 01:24:17.139 over that common denominator. Now add together my fractions on the right side, so they have 01:24:17.140 --> 01:24:18.950 a common denominator. 01:24:18.949 --> 01:24:27.189 So this is x plus three times x plus x plus three. And now I have two fractions that have 01:24:27.189 --> 01:24:32.939 that are equal that have the same denominator, therefore their numerators have to be equal 01:24:32.939 --> 01:24:37.759 also. So the next step is to set the numerators equal. 01:24:37.760 --> 01:24:46.860 So I get x squared is x plus three times x plus x plus three. And if you look back at 01:24:46.859 --> 01:24:52.079 the previous way, we solve this equation, you'll recognize this equation. And so from 01:24:52.079 --> 01:24:57.809 here on we just continue as before. 01:24:57.810 --> 01:25:00.020 When choosing between these two methods, I personally tend 01:25:00.020 --> 01:25:03.600 prefer the clear the denominators method, because it's a little bit less writing, you 01:25:03.600 --> 01:25:07.000 don't have to get rid of those denominators earlier, you don't have to write them as many 01:25:07.000 --> 01:25:12.569 times. But some people find this one a little bit easier to remember, a little easier to 01:25:12.569 --> 01:25:15.359 understand either of these methods is fine. 01:25:15.359 --> 01:25:21.529 One last caution, don't forget at the end, to check your solutions and eliminate any 01:25:21.529 --> 01:25:23.460 extraneous solutions. 01:25:23.460 --> 01:25:30.670 These will be solutions that make the denominators of your original equations go to zero. 01:25:30.670 --> 01:25:36.250 This video is about solving radical equations, that is equations like this one that have 01:25:36.250 --> 01:25:40.789 square root signs in them, or cube roots or any other kind of radical. 01:25:40.789 --> 01:25:45.621 When I see an equation with a square root in it, I really want to get rid of the square 01:25:45.621 --> 01:25:51.079 root. But it'll be easiest to get rid of the square root. If I first isolate the square 01:25:51.079 --> 01:25:56.750 root. In other words, I want to get the term with the square root and that on one side 01:25:56.750 --> 01:26:01.739 of the equation by itself, and everything else on the other side of the equation. If 01:26:01.739 --> 01:26:07.489 I start with my original equation, x plus the square root of x equals 12. And I subtract 01:26:07.489 --> 01:26:12.929 x from both sides, then that does isolate the square root term on the left side with 01:26:12.930 --> 01:26:19.850 everything else on the right. Once I've isolated the term with the square root, I want to get 01:26:19.850 --> 01:26:27.079 rid of the square root. And I'll do that by squaring both sides 01:26:27.079 --> 01:26:36.479 of my equation. So I'll take the square root of x equals 12 minus x and square both sides. 01:26:36.479 --> 01:26:42.309 Now the square root of x squared is just x, taking the square root and then squaring those 01:26:42.310 --> 01:26:45.370 operations undo each other 01:26:45.369 --> 01:26:54.170 to work out 12 minus x squared, write it out and distribute 12 times 12 is 144 12 times 01:26:54.170 --> 01:27:02.539 minus x is minus 12x. I get another minus 12x from here. 01:27:02.539 --> 01:27:09.869 And finally minus x times minus x is positive x squared. So I can combine my minus 12 x's, 01:27:09.869 --> 01:27:18.640 that's minus 24x. And now I can subtract x from both sides to get zero equals 144 minus 01:27:18.640 --> 01:27:26.550 25x plus x squared. That's a quadratic equation, I'll rewrite it in a little bit more standard 01:27:26.550 --> 01:27:27.890 form here. 01:27:27.890 --> 01:27:34.170 So now I've got a familiar quadratic equation with no radical signs left in my equation, 01:27:34.170 --> 01:27:38.800 I'll just proceed to solve it like I usually do for a quadratic, I'll try to factor it. 01:27:38.800 --> 01:27:46.079 So I'm going to look for two numbers that multiply to 144 and add to minus 25. I know 01:27:46.079 --> 01:27:53.079 I'm going to need negative numbers to get to negative 25. And in fact, I'll need two 01:27:53.079 --> 01:27:57.800 negative numbers. So they still multiply to a positive number. So I'll start listing some 01:27:57.800 --> 01:28:04.730 factors of 144, I could have negative one and a negative 144, negative two, and negative 01:28:04.729 --> 01:28:12.079 72, negative four, negative 36, and so on. Once I've listed out the possible factors, 01:28:12.079 --> 01:28:17.780 it's not hard to find the two that add to negative 25. So that's negative nine and negative 01:28:17.780 --> 01:28:28.029 16. So now I can factor in my quadratic equation as x minus nine times x minus 16 equals zero, 01:28:28.029 --> 01:28:34.269 that means that x minus nine is zero or x minus 16 is zero. So x equals nine or x equals 01:28:34.270 --> 01:28:41.000 16. I'm almost done. But there's one last very important step. And that's to check the 01:28:41.000 --> 01:28:49.170 Solutions so that we can eliminate any extraneous solutions and extraneous solution as a solution 01:28:49.170 --> 01:28:54.149 that we get that does not actually satisfy our original equation and extraneous solutions 01:28:54.149 --> 01:29:00.269 can happen when you're solving equations with radicals in them. So let's first check x equals 01:29:00.270 --> 01:29:06.000 nine. If we plug in to our original equation, we get nine plus a squared of nine and we 01:29:06.000 --> 01:29:11.539 want that to equal 12. Well, the square root of nine is three, and nine plus three does 01:29:11.539 --> 01:29:18.390 indeed equal 12. So that solution checks out. Now, let's try x equals 16. Plugging in, we 01:29:18.390 --> 01:29:25.020 get 16 plus a squared of 16. And that's supposed to equal 12. Well, that says 16 plus four 01:29:25.020 --> 01:29:31.880 is supposed to equal 12. But that most definitely is not true. And so x equals 16. Turns out 01:29:31.880 --> 01:29:47.840 to be extraneous solution, and our only solution is x equals nine 01:29:47.840 --> 01:30:26.100 This next equation might not look like an equation involving radicals. But in fact, 01:30:26.100 --> 01:30:32.210 we can think of a fractional exponent as being a radical in disguise. Let's start by doing 01:30:32.210 --> 01:30:38.829 the same thing we did on the previous problem by isolating This time, we'll isolate the 01:30:38.829 --> 01:30:41.590 part of the equation 01:30:41.590 --> 01:30:46.730 that involves the fractional exponent. So I'll start with the original equation, two 01:30:46.729 --> 01:30:53.299 times P to the four fifths equals 1/8. And I'll divide both sides by two or equivalently, 01:30:53.300 --> 01:31:00.000 I can multiply both sides by one half, that gives me p to the four fifths equals 1/16. 01:31:00.000 --> 01:31:05.739 And I've effectively isolated the part of the equation with the fractional exponent 01:31:05.739 --> 01:31:12.920 as much as possible. Now, in the previous example, the next step was to get rid of the 01:31:12.920 --> 01:31:18.980 radical. In this example, we're going to get rid of the fractional exponent. And I'm going 01:31:18.979 --> 01:31:24.799 to actually do this in two stages. First, I'm going to raise both sides to the fifth 01:31:24.800 --> 01:31:26.690 power. 01:31:26.689 --> 01:31:34.509 That's because when I take an exponent to an exponent, I'm multiply my exponents. And 01:31:34.510 --> 01:31:41.840 so that becomes just p to the fourth equals 1/16 to the fifth power. Now, I'm going to 01:31:41.840 --> 01:31:48.360 get rid of the fourth power by raising both sides to the 1/4 power, or by taking the fourth 01:31:48.359 --> 01:31:53.299 root, there's something that you need to be careful about though, when taking 01:31:53.300 --> 01:32:01.880 an even root, or the one over an even number power, you always have to consider plus or 01:32:01.880 --> 01:32:04.630 minus your answer. 01:32:04.630 --> 01:32:10.869 It's kind of like when you write x squared equals four, and you take the square root 01:32:10.869 --> 01:32:16.449 of both sides, x could equal plus or minus the square root of four, right, x could equal 01:32:16.449 --> 01:32:23.659 plus or minus two, since minus two squared is four, just as well as two squared. So that's 01:32:23.659 --> 01:32:32.800 why when you take an even root, or a one over an even power of both sides, you always need 01:32:32.800 --> 01:32:39.440 to include the plus or minus sign, when it's an odd root or one over an odd power, you 01:32:39.439 --> 01:32:46.159 don't need to do that. If you had something like x cubed equals negative eight, then x 01:32:46.159 --> 01:32:51.180 equals the cube root of negative eight, which is negative two would be your only solution, 01:32:51.180 --> 01:32:55.560 you don't need to do the plus or minus because positive two wouldn't work. 01:32:55.560 --> 01:33:02.390 So that aside, explains why we need this plus or minus power. And now p to the four to the 01:33:02.390 --> 01:33:09.590 1/4. When I raise a power to a power, I multiply by exponents, so that's just p to the one, 01:33:09.590 --> 01:33:17.630 which is equal to plus or minus 1/16 to the fifth power to the 1/4 power. 01:33:17.630 --> 01:33:23.010 Now I just need to simplify this expression, I don't really want to raise 1/16 to the fifth 01:33:23.010 --> 01:33:28.760 power, because 16 to the fifth power is like a really huge number. So I think I'm actually 01:33:28.760 --> 01:33:32.420 going to rewrite this first 01:33:32.420 --> 01:33:42.649 as P equals plus or minus 1/16. I'll write it back as the 5/4 power again. 01:33:42.649 --> 01:33:49.239 And as I continue to solve using my exponent rules, 01:33:49.239 --> 01:33:58.939 I'm going to prefer to write this as 1/16 to the fourth root to the fifth power, because 01:33:58.939 --> 01:34:03.649 it's going to be easier to take the fourth root, let's see the fourth root of 1/16 is 01:34:03.649 --> 01:34:08.149 the same thing as the fourth root of one over the fourth root of 16. Raise all that to the 01:34:08.149 --> 01:34:14.359 fifth power. fourth root of one is just one and the fourth root of 16 is to raise that 01:34:14.359 --> 01:34:19.279 to the fifth power, that's just going to be one to the fifth over two to the fifth, which 01:34:19.279 --> 01:34:24.819 is plus or minus 130 seconds. 01:34:24.819 --> 01:34:28.949 The last step is to check answers. 01:34:28.949 --> 01:34:35.880 So I have the two answers p equals 130 seconds, and P equals minus 130 seconds. And if I plug 01:34:35.880 --> 01:34:36.880 those both in 01:34:36.880 --> 01:34:43.199 1/32 to the fourth fifth power. 01:34:43.199 --> 01:34:51.109 That gives me two times one to the fourth power over 32 to the fourth power, which is 01:34:51.109 --> 01:34:59.659 two times one over 32/5 routed to the fourth power. fifth root of 32 is two 01:34:59.659 --> 01:35:07.519 Raisa to the fourth power, I get 16. So this is two times 1/16, which is 1/8, just as we 01:35:07.520 --> 01:35:16.910 wanted in the original equation up here. Similarly, we can check that the P equals negative 1/32 01:35:16.909 --> 01:35:24.039 actually does satisfy the equation, I'll leave that step to the viewer. 01:35:24.039 --> 01:35:30.710 So our two solutions are p equals one over 32, and P equals minus one over 32. I do want 01:35:30.710 --> 01:35:35.310 to point out an alternate approach to getting rid of the fractional exponent, we could have 01:35:35.310 --> 01:35:41.270 gotten rid of it all in one fell swoop by raising both sides of our equation to the 01:35:41.270 --> 01:35:45.340 five fourths power. 01:35:45.340 --> 01:35:51.199 five fourths is the reciprocal of four fifths. So when I use my exponent rules, and say that 01:35:51.199 --> 01:35:56.859 when I raise the power to the power, I multiply my exponents, that gives me p to the four 01:35:56.859 --> 01:36:03.670 fifths times five fourths is plus or minus 1/16 to the five fourths, in other words, 01:36:03.670 --> 01:36:10.569 P to the One Power, which is just P is plus or minus 1/16, to the five fourths, so that's 01:36:10.569 --> 01:36:16.759 an alternate and possibly faster way to get the solution. Once again, the plus or minus 01:36:16.760 --> 01:36:22.440 comes from the fact that when we take the 5/4 power, we're really taking an even root 01:36:22.439 --> 01:36:27.859 a fourth root, and so we need to consider both positive and negative answers. 01:36:27.859 --> 01:36:33.449 This video is about solving radical equations, that is equations like this one that have 01:36:33.449 --> 01:36:39.059 square root signs in them, or cube roots or any other kind of radical. 01:36:39.060 --> 01:36:45.680 In this video, we solved radical equations by first isolating the radical sign, or the 01:36:45.680 --> 01:36:48.270 fractional exponent, 01:36:48.270 --> 01:36:54.670 and then removing the radical sine or the fractional exponent 01:36:54.670 --> 01:37:01.630 by either squaring both sides or taking the reciprocal power of both sides. 01:37:01.630 --> 01:37:06.550 This video is about solving equations with absolute values in them. 01:37:06.550 --> 01:37:12.130 Recall that the absolute value of a positive number is just the number, but the absolute 01:37:12.130 --> 01:37:17.761 value of a negative number is its opposite. In general, I think of the absolute value 01:37:17.761 --> 01:37:23.170 of a number as representing its distance from zero on the number line, 01:37:23.170 --> 01:37:25.840 the number four, 01:37:25.840 --> 01:37:32.360 and the number of negative four are both at distance for from zero, and so the absolute 01:37:32.359 --> 01:37:35.939 value of both of them is four. 01:37:35.939 --> 01:37:44.339 Similarly, if I write the equation, the absolute value of x is three, that means that x has 01:37:44.340 --> 01:37:50.319 to be three units away from zero on the number line. 01:37:50.319 --> 01:37:56.319 And so X would have to be either negative three, or three. Let's start with the equation 01:37:56.319 --> 01:38:00.439 three times the absolute value of x plus two equals four. 01:38:00.439 --> 01:38:08.639 I'd like to isolate the absolute value part of the equation, I can do this by starting 01:38:08.640 --> 01:38:14.090 with my original equation, 01:38:14.090 --> 01:38:17.670 subtracting two from both sides 01:38:17.670 --> 01:38:23.920 and dividing both sides by three. 01:38:23.920 --> 01:38:31.100 Now I'll think in terms of distance on a number line, the absolute value of x is two thirds 01:38:31.100 --> 01:38:40.671 means that x is two thirds away from zero. So x could be here for here at negative two 01:38:40.671 --> 01:38:43.289 thirds or two thirds. 01:38:43.289 --> 01:38:50.420 And the answer to my equation is x is negative two thirds, or two thirds, 01:38:50.420 --> 01:38:52.711 I can check my answers by plugging in 01:38:52.711 --> 01:39:00.090 three times the added value of negative two thirds plus two, I need to check it that equals 01:39:00.090 --> 01:39:04.739 four. Well, the absolute value of negative two thirds is just two thirds. So this is 01:39:04.739 --> 01:39:10.969 three times two thirds plus two, which works out to four. 01:39:10.970 --> 01:39:20.050 Similarly, if I plug in positive two thirds, it also works out to give me the correct answer. 01:39:20.050 --> 01:39:24.239 The second example is a little different, because the opposite value sign is around 01:39:24.239 --> 01:39:29.210 a more complicated expression, not just around the X. 01:39:29.210 --> 01:39:35.930 I would start by isolating the absolute value part. 01:39:35.930 --> 01:39:41.730 But it's already isolated. So I'll just go ahead and jump to thinking about distance 01:39:41.729 --> 01:39:49.899 on the number line. So on my number line, the whole expression 3x plus two is supposed 01:39:49.899 --> 01:39:53.979 to be at a distance of four from zero. 01:39:53.979 --> 01:39:59.939 So that means that 3x plus two is here at four or 3x plus two 01:39:59.939 --> 01:40:03.839 though is it negative four, all right, those as equations 01:40:03.840 --> 01:40:12.090 3x plus two equals four, or 3x plus two is minus four. And then I can solve. 01:40:12.090 --> 01:40:20.369 So this becomes 3x equals two, or x equals two thirds. And over here, I get 3x equals 01:40:20.369 --> 01:40:29.489 minus six, or x equals minus two. Finally, I'll check my answers. 01:40:29.489 --> 01:40:32.800 I'll leave it to you to verify that they both work. 01:40:32.800 --> 01:40:38.590 A common mistake on absolute value equations is to get rid of the absolute value signs 01:40:38.590 --> 01:40:44.569 like we did here, and then just solve for one answer, instead of solving for both answers. 01:40:44.569 --> 01:40:48.759 Another mistake sometimes people make is, once they get the first answer, they just 01:40:48.760 --> 01:40:52.640 assume that the negative of that works also. 01:40:52.640 --> 01:40:57.900 But that doesn't always work. In the first example, our two answers were both the negatives 01:40:57.899 --> 01:41:02.710 of each other. But in our second examples, or two answers, were not just the opposites 01:41:02.710 --> 01:41:06.619 of each other one was two thirds and the other was negative two. 01:41:06.619 --> 01:41:12.659 In this third example, let's again, isolate the absolute value part of the equation. 01:41:12.659 --> 01:41:16.760 So starting with our original equation, 01:41:16.760 --> 01:41:22.550 we can subtract 16 from both sides, 01:41:22.550 --> 01:41:31.320 and divide both sides by five or equivalently, multiply by 1/5. 01:41:31.319 --> 01:41:35.049 Now let's think about distance on the number line, 01:41:35.050 --> 01:41:41.770 we have an absolute value needs to equal negative three. So that means whatever is inside the 01:41:41.770 --> 01:41:48.400 absolute value sign needs to be at distance negative three from zero, well, you can't 01:41:48.399 --> 01:41:52.849 be at distance negative three from zero. Another way of thinking about this is you can't have 01:41:52.850 --> 01:41:57.220 the absolute value of something and end up with a negative number if the value is always 01:41:57.220 --> 01:42:05.530 positive, or zero. So this equation doesn't actually make sense, and there are no solutions 01:42:05.529 --> 01:42:07.509 to this equation. 01:42:07.510 --> 01:42:15.010 In this video, we solved absolute value equations. In many cases, an absolute value equation 01:42:15.010 --> 01:42:20.949 will have two solutions. But in some cases, it'll have no solutions. And occasionally, 01:42:20.949 --> 01:42:25.079 it'll have just one solution. 01:42:25.079 --> 01:42:30.979 This video is about interval notation, and easy and well known way to record inequalities. 01:42:30.979 --> 01:42:39.159 Before dealing with interval notation, it is important to know how to deal with inequalities. 01:42:39.159 --> 01:42:45.420 Our first example of an inequality is written here, all numbers between one and three, not 01:42:45.420 --> 01:42:46.750 including one and three. 01:42:46.750 --> 01:42:53.260 First, we used to write down our variable x, that it is important to locate the key 01:42:53.260 --> 01:43:01.320 values for this problem that is one and three. Now here we see that x is between these two 01:43:01.319 --> 01:43:07.529 numbers, meaning we will have one inequality statement on each side of the variable. Here 01:43:07.529 --> 01:43:12.779 it says not including one and three, which means that we do not have an or equal to sign 01:43:12.779 --> 01:43:18.670 beneath each inequality. Here we will put one the lowest key value, and here we will 01:43:18.670 --> 01:43:24.489 put three the highest key value. Next, we're going to graph this inequality on a number 01:43:24.489 --> 01:43:27.069 line. 01:43:27.069 --> 01:43:34.029 Here we write our key values one, and three. Because it is not including one, and three, 01:43:34.029 --> 01:43:40.189 we have an empty circle around each number. Because it isn't the numbers between we have 01:43:40.189 --> 01:43:41.289 a line connecting them. 01:43:41.289 --> 01:43:50.069 The last step of this problem is writing this inequality in interval notation. 01:43:50.069 --> 01:43:54.949 writing things in interval notation is kind of like writing an ordered pair. Here we will 01:43:54.949 --> 01:43:57.579 put one and then here we will put three. 01:43:57.579 --> 01:44:03.189 Next we need to put brackets around these numbers. 01:44:03.189 --> 01:44:09.429 For this problem, it is not including one in three, which means we will use soft brackets. 01:44:09.430 --> 01:44:15.970 However, if it were including one and three, we will use hard brackets. 01:44:15.970 --> 01:44:22.960 It is important to note for interval notation, that the smallest value always goes on the 01:44:22.960 --> 01:44:29.039 left and the biggest value always goes on the right. 01:44:29.039 --> 01:44:32.680 You also include a comma between your two key values. 01:44:32.680 --> 01:44:41.770 Now let's work on problem B. Once again we write our variable x. Again it is between 01:44:41.770 --> 01:44:47.170 negative four and two but this time it is including a negative four into this requires 01:44:47.170 --> 01:44:53.859 us to have the or equal to assign below the inequality. Then we put our lowest key value 01:44:53.859 --> 01:44:59.739 here and their highest here. The number line graph for this problem is slightly different. 01:44:59.739 --> 01:45:00.739 We still 01:45:00.739 --> 01:45:07.399 have our key values negative four and two. But instead of an open circle, like we have 01:45:07.399 --> 01:45:15.759 right here, we instead use a closed circle, representing that it is including negative 01:45:15.760 --> 01:45:21.320 four and two, you complete this with a line in between. Last we write this in interval 01:45:21.319 --> 01:45:27.170 notation. In the last problem, I said that for numbers, including the outside values, 01:45:27.170 --> 01:45:35.470 we would use these hard brackets. Now we are actually using this. So we have or hard bracket 01:45:35.470 --> 01:45:40.650 on each side because isn't including foreign zoo, then we put our smallest value on the 01:45:40.649 --> 01:45:46.799 left, and our highest value on the right, with the comma in between. 01:45:46.800 --> 01:45:51.029 You now know how to correctly write these two types of intervals. 01:45:51.029 --> 01:45:58.689 Now we are going to practice transforming these from inequality notation to interval 01:45:58.689 --> 01:46:05.569 notation, and vice versa. This is slightly more difficult than our previous examples 01:46:05.569 --> 01:46:11.939 of having two soft brackets or two hard brackets, we'll have two different types. Now on the 01:46:11.939 --> 01:46:18.559 left side will have a hard bracket because it is including the three. However, on the 01:46:18.560 --> 01:46:24.039 right side, it is a soft bracket because it is it it is not including the one that we 01:46:24.039 --> 01:46:31.880 put a comma in the middle, a lower key value and our higher key value. For the next problem, 01:46:31.880 --> 01:46:37.140 we're taking an equation already written an interval notation and putting it back into 01:46:37.140 --> 01:46:38.430 inequality notation. 01:46:38.430 --> 01:46:47.600 For the second problem, we can see are key values as being five and negative infinity. 01:46:47.600 --> 01:46:54.340 Now, this sounds a little bit weird, but let's just set up or an equality, we have our variable 01:46:54.340 --> 01:47:02.470 x, which is less than because of the soft bracket five, and greater than without the 01:47:02.470 --> 01:47:09.409 or equal to because it has a soft bracket there to infinity. But because x is always 01:47:09.409 --> 01:47:16.319 greater than negative infinity, we can take out this part, leaving us with x is less than 01:47:16.319 --> 01:47:17.739 five. 01:47:17.739 --> 01:47:25.269 It is important to note that a soft bracket always accompanies infinity. This is because 01:47:25.270 --> 01:47:32.010 infinity is not a real number, so we cannot include it just go as far up to it as we can. 01:47:32.010 --> 01:47:38.110 The next problem is a little tricky, because we don't see another key value over here. 01:47:38.109 --> 01:47:43.829 But let's just start off the equation as a soft bracket on this side due to the absence 01:47:43.829 --> 01:47:51.510 of an order equal to sign. And negative 15 is the lower key value. 01:47:51.510 --> 01:47:57.190 On the other side, we put the highest possible number infinity, and then close it off with 01:47:57.189 --> 01:48:05.289 a soft bracket. We can see the relation of this to the previous problem. And how it goes 01:48:05.289 --> 01:48:12.119 into this problem when x is greater than a key value instead of less than a key value. 01:48:12.119 --> 01:48:19.189 Once again, we have our sauce bracket for infinity, which is always true. 01:48:19.189 --> 01:48:26.279 Part D brings up an important point about which number goes on the left. For all of 01:48:26.279 --> 01:48:30.130 our other problems, we have had the inequalities pointing left 01:48:30.130 --> 01:48:41.609 instead of rights. When seeing Part D, you might think oh, four is on the left, so it 01:48:41.609 --> 01:48:47.799 would go here, and zero is on the right who would go here. But that is not true. 01:48:47.800 --> 01:48:53.940 When writing inequalities, you must always have the lower value on the left, which for 01:48:53.939 --> 01:49:01.550 this problem is zero. To fix this, we can simply flip this inequality. So it reads like 01:49:01.551 --> 01:49:06.800 this, it is still identical just written in an easier form to transform it into interval 01:49:06.800 --> 01:49:13.070 notation, then we can see that zero has a soft bracket, because it is not including 01:49:13.069 --> 01:49:19.899 zero, but r comma four or other key value and a hard backup because it is or equal to 01:49:19.899 --> 01:49:21.859 four. 01:49:21.859 --> 01:49:29.420 For interval notation, you must always have the smaller number on the left side. 01:49:29.420 --> 01:49:36.190 This was our video on interval notation, an alternative way to write inequalities. 01:49:36.189 --> 01:49:41.739 This video is about solving inequalities that have absolute value signs in them. 01:49:41.739 --> 01:49:45.979 Let's look at the inequality absolute value of x is less than five on the number line. 01:49:45.979 --> 01:49:52.529 Thinking of absolute value as distance. This means that the distance between x and zero 01:49:52.529 --> 01:50:01.380 is less than five units. So x has to live somewhere in between negative five and five 01:50:01.380 --> 01:50:06.560 We can express this as an inequality without absolute value signs by saying negative five 01:50:06.560 --> 01:50:12.850 is less than x, which is less than five. Or we can use interval notation, soft bracket 01:50:12.850 --> 01:50:16.890 negative five, five, soft bracket. 01:50:16.890 --> 01:50:20.760 Both of these formulations are equivalent to the original one, but don't involve the 01:50:20.760 --> 01:50:23.650 absolute value signs. 01:50:23.649 --> 01:50:28.670 In the second example, we're looking for the values of x for which the absolute value of 01:50:28.670 --> 01:50:32.649 x is greater than or equal to five. 01:50:32.649 --> 01:50:39.299 on the number line, this means that the distance of x from the zero, it's got to be bigger 01:50:39.300 --> 01:50:42.070 than or equal to five units. 01:50:42.069 --> 01:50:48.130 A distance bigger than five units means that x has to live somewhere over here, or somewhere 01:50:48.130 --> 01:50:54.730 over here, where it's farther than five units away from zero. Course x could also have a 01:50:54.729 --> 01:51:00.320 distance equal to five units. So I'll fill in that dot and shade in the other parts of 01:51:00.320 --> 01:51:04.569 the number line that satisfy my inequality. 01:51:04.569 --> 01:51:10.130 Now I can rewrite the inequality without the absolute value symbols by saying that x is 01:51:10.130 --> 01:51:17.090 less than or equal to negative five, or x is greater than or equal to five. I could 01:51:17.090 --> 01:51:20.239 also write this in interval notation, 01:51:20.239 --> 01:51:26.689 soft bracket, negative infinity, negative five, hard bracket. And the second part is 01:51:26.689 --> 01:51:33.829 hard bracket five infinity soft bracket, I combine these with a u for union. Because 01:51:33.829 --> 01:51:38.269 I'm trying to describe all these points on the number line together with all these other 01:51:38.270 --> 01:51:39.920 points. 01:51:39.920 --> 01:51:44.520 Let's take this analysis a step further with a slightly more complicated problem. Now I 01:51:44.520 --> 01:51:49.010 want the absolute value of three minus two t to be less than four. 01:51:49.010 --> 01:51:56.020 And an absolute value less than four means a distance less than four on the number line. 01:51:56.020 --> 01:52:00.260 But it's not the variable t that lives in here at a distance of less than four from 01:52:00.260 --> 01:52:04.960 zero, it's the whole expression, three minus two t. 01:52:04.960 --> 01:52:13.980 So three minus two t, live somewhere in here. And I can rewrite this as an inequality without 01:52:13.979 --> 01:52:21.209 absolute value signs by saying negative four is less than three minus two t is less than 01:52:21.210 --> 01:52:27.730 four. Now I have a compound inequality that I can solve the usual way. First, I subtract 01:52:27.729 --> 01:52:33.739 three from all three sides to get negative seven is less than negative two t is less 01:52:33.739 --> 01:52:40.719 than one. And now I'll divide all three sides by negative two. Since negative two is a negative 01:52:40.720 --> 01:52:46.140 number, this reverses the directions of the inequalities. 01:52:46.140 --> 01:52:52.050 Simplifying, I get seven halves is greater than t is greater than negative one half. 01:52:52.050 --> 01:52:58.720 So my final answer on the number line looks like all the stuff between negative a half 01:52:58.720 --> 01:53:01.880 and seven halves. 01:53:01.880 --> 01:53:07.340 But not including the endpoints, and an interval notation, I can write this soft bracket negative 01:53:07.340 --> 01:53:09.920 a half, seven, half soft bracket, 01:53:09.920 --> 01:53:14.539 please pause the video and try the next problem on your own. 01:53:14.539 --> 01:53:19.909 thinking in terms of distance, this inequality says that the distance between the expression 01:53:19.909 --> 01:53:25.880 three minus two t and zero is always bigger than four. Let me draw this on the number 01:53:25.880 --> 01:53:27.609 line. 01:53:27.609 --> 01:53:33.779 If three minus two t has a distance bigger than four from zero, then it can't be in this 01:53:33.779 --> 01:53:39.869 region that's near zero, it has to be on the outside in one of these two regions. 01:53:39.869 --> 01:53:48.569 That is three minus two t is either less than negative four, or three minus two t is bigger 01:53:48.569 --> 01:53:54.369 than four. I solve these two inequalities separately, the first one, subtracting three 01:53:54.369 --> 01:54:01.390 from both sides, I get negative two t is less than negative seven divided by negative two, 01:54:01.390 --> 01:54:08.350 I get T is bigger than seven halves. And then on the other side, I get negative two t is 01:54:08.350 --> 01:54:15.820 greater than one. So t is less than negative one half, I had to flip inequalities in the 01:54:15.819 --> 01:54:21.259 last step due to dividing by a negative number. So let's check this out on the number line 01:54:21.260 --> 01:54:28.750 again, the first piece says that t is greater than seven halves. I'll draw that over here. 01:54:28.750 --> 01:54:41.029 And the second piece says that t is less than negative one half. I'll draw that down here. 01:54:41.029 --> 01:54:46.380 Because these two statements are joined with an or I'm looking for the t values that are 01:54:46.380 --> 01:54:52.690 in this one, or in this one. That is I want both of these regions put together. So an 01:54:52.689 --> 01:55:00.029 interval notation, this reads negative infinity to negative one half soft bracket union soft 01:55:00.029 --> 01:55:01.029 bracket 01:55:01.029 --> 01:55:03.130 Seven halves to infinity. 01:55:03.130 --> 01:55:09.289 This last example looks more complicated. But if I simplify first and isolate the absolute 01:55:09.289 --> 01:55:14.500 value part, it looks pretty much like the previous ones. So I'll start by subtracting 01:55:14.500 --> 01:55:20.310 seven from both sides. And then I'll divide both sides by two. Now I'll draw my number 01:55:20.310 --> 01:55:21.310 line. 01:55:21.310 --> 01:55:28.060 And I'm looking for this expression for x plus five, to always have distance greater 01:55:28.060 --> 01:55:33.471 than or equal to negative three from zero, wait a second, distance greater than equal 01:55:33.470 --> 01:55:36.759 to negative three, well, distance is always greater than or equal to negative three is 01:55:36.760 --> 01:55:43.340 always greater than equal to zero. So this, in fact, is always true. 01:55:43.340 --> 01:55:50.119 And so the answer to my inequality is all numbers between negative infinity and infinity. 01:55:50.119 --> 01:55:55.930 In other words, all real numbers. 01:55:55.930 --> 01:56:01.400 Once solving absolute value inequalities, it's good to think about distance. And absolute 01:56:01.399 --> 01:56:08.500 value of something that's less than a number means that whatever's inside the absolute 01:56:08.500 --> 01:56:12.869 value signs is close to zero. 01:56:12.869 --> 01:56:18.579 On the other hand, an absolute value is something and being greater than a number means that 01:56:18.579 --> 01:56:25.829 whatever's inside the absolute value sign is far away from zero, because its distance 01:56:25.829 --> 01:56:29.960 from zero is bigger than that certain number. 01:56:29.960 --> 01:56:34.739 drawing these pictures on the number line is a helpful way to rewrite the absolute value 01:56:34.739 --> 01:56:40.899 and equality as an inequality that doesn't contain an absolute value sign. In this case, 01:56:40.899 --> 01:56:47.179 it would be negative three is less than x plus two is less than three. And in the other 01:56:47.180 --> 01:56:53.630 case, it would be either x plus two is less than negative three, or x plus two is greater 01:56:53.630 --> 01:56:55.880 than three. 01:56:55.880 --> 01:57:01.670 This video is about solving linear inequalities. Those are inequalities like this one that 01:57:01.670 --> 01:57:09.170 involve a variable here x, but don't involve any x squared or other higher power terms. 01:57:09.170 --> 01:57:15.739 The good news is, we can solve linear inequalities, just like we solve linear equations by distributing, 01:57:15.739 --> 01:57:19.779 adding and subtracting terms to both sides and multiplying and dividing by numbers on 01:57:19.779 --> 01:57:26.719 both sides. The only thing that's different is that if you multiply or divide by a negative 01:57:26.720 --> 01:57:32.840 number, then you need to reverse the direction of the inequality. For example, if we had 01:57:32.840 --> 01:57:39.319 the inequality, negative x is less than negative five, and we wanted to multiply both sides 01:57:39.319 --> 01:57:45.710 by negative one to get rid of the negative signs, we'd have to also switch or reverse 01:57:45.710 --> 01:57:50.560 the inequality. With this caution in mind, let's look at our first example. 01:57:50.560 --> 01:57:56.170 Since our variable x is trapped in parentheses, I'll distribute the negative five to free 01:57:56.170 --> 01:57:58.569 it from the parentheses. 01:57:58.569 --> 01:58:04.849 That gives me negative 5x minus 10 plus three is greater than eight. 01:58:04.850 --> 01:58:10.020 Negative 10 plus three is negative seven, so I'll rewrite this as negative 5x minus 01:58:10.020 --> 01:58:16.890 seven is greater than eight. Now I'll add seven to both sides to get negative 5x is 01:58:16.890 --> 01:58:23.100 greater than 15. Now I'd like to divide both sides by negative five. Since negative five 01:58:23.100 --> 01:58:31.260 is a negative number that reverses the inequality. So I get x is less than 15 divided by negative 01:58:31.260 --> 01:58:36.130 five. In other words, x is less than negative three. 01:58:36.130 --> 01:58:40.390 If I wanted to graph this on a number line, I just need to put down a negative three, 01:58:40.390 --> 01:58:44.400 an open circle around it, and shade in to the left, 01:58:44.399 --> 01:58:51.049 I use an open circle, because x is strictly less than negative three and can't equal negative 01:58:51.050 --> 01:58:56.070 three. If I wanted to write this in interval notation, I've added a soft bracket negative 01:58:56.069 --> 01:59:01.989 infinity, negative three soft bracket. Again, the soft bracket is because the negative three 01:59:01.989 --> 01:59:10.090 is not included. This next example is an example of a compound inequality. It has two parts. 01:59:10.090 --> 01:59:15.900 Either this statement is true, or this statement is true. I want to find the values of x that 01:59:15.899 --> 01:59:21.529 satisfy either one. I'll solve this by working out each part separately, and then combining 01:59:21.529 --> 01:59:28.079 them at the end. For the inequality on the left, I'll copy it over here. I'm going to 01:59:28.079 --> 01:59:32.149 add four to both sides. 01:59:32.149 --> 01:59:36.189 then subtract x from both sides 01:59:36.189 --> 01:59:40.149 and then divide both sides by two. 01:59:40.149 --> 01:59:45.170 I didn't have to reverse the inequality sign because I divided by a positive number 01:59:45.170 --> 01:59:51.640 on the right side, I'll copy the equation over, subtract one from both sides, and divide 01:59:51.640 --> 01:59:59.329 both sides by 696 is the same as three halves. Now I'm looking for the x values that make 01:59:59.329 --> 02:00:00.329 this statement. 02:00:00.329 --> 02:00:07.300 row four, make this statement true. Let me graph this on a number line. 02:00:07.300 --> 02:00:14.329 x is less than or equal to negative two, means I put a filled in circle there and graph everything 02:00:14.329 --> 02:00:22.670 to the left. X is greater than three halves means I put a empty circle here and shaded 02:00:22.670 --> 02:00:25.239 and everything to the right. 02:00:25.239 --> 02:00:33.050 My final answer includes both of these pieces, which I'll reshade in green, because x is 02:00:33.050 --> 02:00:38.880 allowed to be an either one or the other. Finally, I can write this in interval notation. 02:00:38.880 --> 02:00:44.340 The first piece on the number line can be described as soft bracket negative infinity, 02:00:44.340 --> 02:00:50.900 negative two hard bracket. And the second piece can be described as soft bracket three 02:00:50.899 --> 02:00:57.379 halves, infinity soft bracket to indicate that x can be in either one of these pieces, 02:00:57.380 --> 02:01:01.590 I use the union side, which is a U. 02:01:01.590 --> 02:01:09.340 That means that my answer includes all x values in here together with all x values in here. 02:01:09.340 --> 02:01:15.079 This next example is also a compound inequality. This time I'm joined by an ad, the and means 02:01:15.079 --> 02:01:22.029 I'm looking for all y values that satisfy both this and this at the same time. Again, 02:01:22.029 --> 02:01:25.340 I can solve each piece separately 02:01:25.340 --> 02:01:34.090 on the left, to isolate the Y, I need to multiply by negative three halves on both sides. So 02:01:34.090 --> 02:01:40.460 that gives me Why is less than negative 12 times negative three halves, the greater than 02:01:40.460 --> 02:01:46.649 flip to a less than because I was multiplying by negative three halves, which is a negative 02:01:46.649 --> 02:01:47.729 number. 02:01:47.729 --> 02:01:54.469 By clean up the right side, I get why is less than 18. 02:01:54.470 --> 02:01:59.590 On the right side, I'll start by subtracting two from both sides. 02:01:59.590 --> 02:02:07.300 And now I'll divide by negative four, again, a negative number, so that flips the inequality. 02:02:07.300 --> 02:02:13.270 So that's why is less than three over negative four. In other words, y is less than negative 02:02:13.270 --> 02:02:19.710 three fourths. Again, I'm looking for the y values that make both of these statements 02:02:19.710 --> 02:02:22.230 true at the same time. 02:02:22.229 --> 02:02:25.819 Let me graph this on the number line, 02:02:25.819 --> 02:02:34.819 the Y is less than 18. I can graph that by drawing the number 18. 02:02:34.819 --> 02:02:41.519 I don't want to include it. So I use an empty circle and I shaded everything to the left 02:02:41.520 --> 02:02:47.490 the statement y is less than negative three fourths, I need to draw a negative three fourths. 02:02:47.489 --> 02:02:54.529 And again, I don't include it, but I do include everything on the left. Since I want the y 02:02:54.529 --> 02:02:59.380 values for which both of these statements are true, I need the y values that are in 02:02:59.380 --> 02:03:05.900 both colored blue and colored red. And so that would be this part right here. I'll just 02:03:05.899 --> 02:03:10.819 draw it above so you can see it easily. So that would be all the numbers from negative 02:03:10.819 --> 02:03:15.869 three fourths and lower, not including negative three, four, since those are the parts of 02:03:15.869 --> 02:03:21.380 the number line that have both red and blue colors on them. In interval notation, my final 02:03:21.380 --> 02:03:26.920 answer will be soft bracket negative infinity to negative three fourths soft bracket. 02:03:26.920 --> 02:03:34.720 As my final example, I have an inequality that has two inequality signs in it negative 02:03:34.720 --> 02:03:40.380 three is less than or equal to 6x minus two is less than 10. I can think of this as being 02:03:40.380 --> 02:03:47.750 a compound inequality with two parts, negative 3x is less than or equal to 6x minus two. 02:03:47.750 --> 02:03:54.289 And at the same time 6x minus two is less than 10. I could solve this into pieces as 02:03:54.289 --> 02:04:00.460 before. But instead, it's a little more efficient to just solve it all at once, by doing the 02:04:00.460 --> 02:04:08.010 same thing to all three sides. So as a first step, I'll add two to all three sides. That 02:04:08.010 --> 02:04:14.400 gives me negative one is less than or equal to 6x is less than 12. And now I'm going to 02:04:14.399 --> 02:04:20.149 divide all three sides by six to isolate the x. So that gives me negative one, six is less 02:04:20.149 --> 02:04:24.189 than or equal to x is less than two. 02:04:24.189 --> 02:04:29.129 If we solved it, instead, in two pieces above, we'd end up with the same thing because we 02:04:29.130 --> 02:04:35.150 get negative one six is less than or equal to x from this piece, and we'd get x is less 02:04:35.149 --> 02:04:40.299 than two on this piece. And because of the and statement, that's the same thing as saying 02:04:40.300 --> 02:04:45.000 negative one, six is less than or equal to x, which is less than two. 02:04:45.000 --> 02:04:50.050 Either way we do it. Let's see if what it looks like on the number line. So on the number 02:04:50.050 --> 02:04:58.829 line, we're looking for things that are between two and negative one six, including the negative 02:04:58.829 --> 02:04:59.970 one, six 02:04:59.970 --> 02:05:05.470 But not including the to interval notation, we can write this as hard bracket, negative 02:05:05.470 --> 02:05:08.510 162 soft bracket. 02:05:08.510 --> 02:05:13.600 In this video, we solve linear inequalities, including some compound inequalities, joined 02:05:13.600 --> 02:05:18.670 by the conjunctions and, and or, 02:05:18.670 --> 02:05:22.789 remember when we're working with and we're looking for places on the number line where 02:05:22.789 --> 02:05:24.899 both statements are true. 02:05:24.899 --> 02:05:30.769 That is, we're looking for the overlap on the number line. 02:05:30.770 --> 02:05:35.510 In this case, the points on the number line that are colored both red and blue at the 02:05:35.510 --> 02:05:36.650 same time. 02:05:36.649 --> 02:05:43.089 When we're working with oral statements, we're looking for places where either one or the 02:05:43.090 --> 02:05:46.380 other statement is true or both 02:05:46.380 --> 02:05:49.010 on the number line, 02:05:49.010 --> 02:05:54.380 this corresponds to points that are colored either red or blue, or both. And in this picture, 02:05:54.380 --> 02:05:57.949 that will actually correspond to the entire number line. 02:05:57.949 --> 02:06:04.630 In this video, we'll solve inequalities involving polynomials like this one, and inequalities 02:06:04.630 --> 02:06:09.340 involving rational expressions like this one. 02:06:09.340 --> 02:06:14.500 Let's start with a simple example. Maybe a deceptively simple example. If you see the 02:06:14.500 --> 02:06:19.550 inequality, x squared is less than four, you might be very tempted to take the square root 02:06:19.550 --> 02:06:26.220 of both sides and get something like x is less than two as your answer. But in fact, 02:06:26.220 --> 02:06:27.789 that doesn't work. 02:06:27.789 --> 02:06:35.850 To see why it's not correct, consider the x value of negative 10. 02:06:35.850 --> 02:06:43.500 Negative 10 satisfies the inequality, x is less than two since negative 10 is less than 02:06:43.500 --> 02:06:51.130 two. But it doesn't satisfy the inequality x squared is less than four, since negative 02:06:51.130 --> 02:06:58.420 10 squared is 100, which is not less than four. So these two inequalities are not the 02:06:58.420 --> 02:07:04.230 same. And it doesn't work to solve a quadratic inequality just to take the square root of 02:07:04.229 --> 02:07:09.349 both sides, you might be thinking part of why this reasoning is wrong, as we've ignored 02:07:09.350 --> 02:07:15.420 the negative two option, right? If we had the equation, x squared equals four, then 02:07:15.420 --> 02:07:21.430 x equals two would just be one option, x equals negative two would be another solution. So 02:07:21.430 --> 02:07:27.869 somehow, our solution to this inequality should take this into account. In fact, a good way 02:07:27.869 --> 02:07:33.960 to solve an inequality involving x squares or higher power terms, is to solve the associated 02:07:33.960 --> 02:07:40.399 equation first. But before we even do that, I like to pull everything over to one side, 02:07:40.399 --> 02:07:44.449 so that my inequality has zero on the other side. 02:07:44.449 --> 02:07:49.569 So for our equation, I'll subtract four from both sides to get x squared minus four is 02:07:49.569 --> 02:07:50.590 less than zero. 02:07:50.590 --> 02:07:58.039 Now, I'm going to actually solve the associated equation, x squared minus four is equal to 02:07:58.039 --> 02:08:06.420 zero, I can do this by factoring to x minus two times x plus two is equal to zero. And 02:08:06.420 --> 02:08:12.579 I'll set my factors equal to zero, and I get x equals two and x equals minus two. 02:08:12.579 --> 02:08:17.970 Now, I'm going to plot the solutions to my equation on the number line. So I write down 02:08:17.970 --> 02:08:23.240 negative two and two, those are the places where my expression x squared minus four is 02:08:23.239 --> 02:08:26.170 equal to zero. 02:08:26.170 --> 02:08:30.550 Since I want to find where x squared minus four is less than zero, I want to know where 02:08:30.550 --> 02:08:36.029 this expression x squared minus four is positive or negative, a good way to find that out is 02:08:36.029 --> 02:08:43.489 to plug in test values. So first, a plug in a test value in this area of the number line, 02:08:43.489 --> 02:08:47.369 something less than negative to say x equals negative three. 02:08:47.369 --> 02:08:53.720 If I plug in negative three into x squared minus four, I get negative three squared minus 02:08:53.720 --> 02:09:02.090 four, which is nine minus four, which is five, that's a positive number. So at negative three, 02:09:02.090 --> 02:09:08.029 the expression x squared minus four is positive. And in fact, everywhere on this region of 02:09:08.029 --> 02:09:12.420 the number line, my expression is going to be positive, because it can jump from positive 02:09:12.420 --> 02:09:17.529 to negative without going through a place where it's zero, I can figure out whether 02:09:17.529 --> 02:09:21.859 x squared minus four is positive or negative on this region, and on this region of the 02:09:21.859 --> 02:09:25.109 number line by plugging in test value similar way, 02:09:25.109 --> 02:09:31.849 evaluate the plug in between negative two and two, a nice value is x equals 00 squared 02:09:31.850 --> 02:09:36.700 minus four, that's negative four and negative number. So I know that my expression x squared 02:09:36.699 --> 02:09:43.109 minus four is negative on this whole interval. Finally, I can plug in something like x equals 02:09:43.109 --> 02:09:49.349 10, something bigger than two, and I get 10 squared minus four. Without even computing 02:09:49.350 --> 02:09:53.890 that I can tell that that's going to be a positive number. And that's all that's important. 02:09:53.890 --> 02:09:57.980 Again, since I want x squared minus four to be less than zero, I'm looking for the places 02:09:57.979 --> 02:09:59.719 on this number line where I'm getting 02:09:59.720 --> 02:10:06.230 negatives. So I will share that in on my number line. It's in here, not including the endpoints. 02:10:06.229 --> 02:10:10.609 Because the endpoints are where my expression x squared minus four is equal to zero, and 02:10:10.609 --> 02:10:12.599 I want it strictly less than zero, 02:10:12.600 --> 02:10:19.900 I can write my answer as an inequality, negative two is less than x is less than two, or an 02:10:19.899 --> 02:10:25.059 interval notation as soft bracket negative two to soft bracket. 02:10:25.060 --> 02:10:32.780 Our next example, we can solve similarly, first, we'll move everything to one side, 02:10:32.779 --> 02:10:39.769 so that our inequality is x cubed minus 5x squared minus 6x is greater than or equal 02:10:39.770 --> 02:10:47.420 to zero. Next, we'll solve the associated equation by factoring. So first, I'll write 02:10:47.420 --> 02:10:55.449 down the equation. Now I'll factor out an x. And now I'll factor the quadratic. So the 02:10:55.449 --> 02:11:01.239 solutions to my equation are x equals 0x equals six and x equals negative one, 02:11:01.239 --> 02:11:05.489 I'll write the solutions to the equation on the number line. 02:11:05.489 --> 02:11:17.359 So that's negative one, zero, and six. That's where my expression x times x minus six times 02:11:17.359 --> 02:11:21.049 x plus one is equal to zero. 02:11:21.050 --> 02:11:28.110 But I want to find where it's greater than or equal to zero. So again, I can use test 02:11:28.109 --> 02:11:35.630 values, I can plug in, for example, x equals negative two, either to this version of expression, 02:11:35.630 --> 02:11:40.760 or to this factored version. Since I only care whether my answer is positive or negative, 02:11:40.760 --> 02:11:45.630 it's sometimes easier to use the factored version. For example, when x is negative two, 02:11:45.630 --> 02:11:48.069 this factor is negative. 02:11:48.069 --> 02:11:55.539 But this factor, x minus six is also negative when I plug in negative two for x. 02:11:55.539 --> 02:12:03.269 Finally, x plus one, when I plug in negative two for x, that's negative one, that's also 02:12:03.270 --> 02:12:08.980 negative. And a negative times a negative times a negative gives me a negative number. 02:12:08.979 --> 02:12:16.750 If I plug in something, between negative one and zero, say x equals negative one half, 02:12:16.750 --> 02:12:21.600 then I'm going to get a negative for this factor, a negative for this factor, but a 02:12:21.600 --> 02:12:25.020 positive for this third factor. 02:12:25.020 --> 02:12:28.860 Negative times negative times positive gives me a positive 02:12:28.859 --> 02:12:34.349 for a test value between zero and six, let's try x equals one. 02:12:34.350 --> 02:12:41.130 Now I'll get a positive for this factor a negative for this factor, and a positive for 02:12:41.130 --> 02:12:43.880 this factor. 02:12:43.880 --> 02:12:50.920 positive times a negative times a positive gives me a negative. Finally, for a test value 02:12:50.920 --> 02:12:56.649 bigger than six, we could use say x equals 100. That's going to give me positive, positive 02:12:56.649 --> 02:12:59.219 positive. 02:12:59.220 --> 02:13:03.630 So my product will be positive. 02:13:03.630 --> 02:13:08.060 Since I want values where my expression is greater than or equal to zero, I want the 02:13:08.060 --> 02:13:12.970 places where it equals zero. 02:13:12.970 --> 02:13:16.980 And the places where it's positive. 02:13:16.979 --> 02:13:24.709 So my final answer will be close bracket negative one to zero, close bracket union, close bracket 02:13:24.710 --> 02:13:25.909 six to infinity. 02:13:25.909 --> 02:13:34.489 As our final example, let's consider the rational inequality, x squared plus 6x plus nine divided 02:13:34.489 --> 02:13:38.229 by x minus one is less than or equal to zero. 02:13:38.229 --> 02:13:43.449 Although it might be tempting to clear the denominator and multiply both sides by x minus 02:13:43.449 --> 02:13:49.050 one, it's dangerous to do that, because x minus one could be a positive number. But 02:13:49.050 --> 02:13:53.699 it could also be a negative number. And when you multiply both sides by a negative number, 02:13:53.699 --> 02:13:59.260 you have to reverse the inequality. Although it's possible to solve the inequality this 02:13:59.260 --> 02:14:04.869 way, by thinking of cases where x minus one is less than zero or bigger than zero, I think 02:14:04.869 --> 02:14:10.369 it's much easier just to solve the same way as we did before. So we'll start by rewriting 02:14:10.369 --> 02:14:15.340 so that we move all terms to the left and have zero on the right. Well, that's already 02:14:15.340 --> 02:14:20.750 true. So the next step would be to solve the associated equation. 02:14:20.750 --> 02:14:29.470 That is x squared plus 6x plus nine over x minus one is equal to zero. 02:14:29.470 --> 02:14:35.240 That would be where the numerator is 0x squared plus 6x plus nine is equal to zero. So we're 02:14:35.239 --> 02:14:42.609 x plus three squared is zero, or x equals negative three. There's one extra step we 02:14:42.609 --> 02:14:48.069 have to do for rational expressions. And that's we need to find where the expression does 02:14:48.069 --> 02:14:55.559 not exist. That is, let's find where the denominator is zero. And that said, x equals one. 02:14:55.560 --> 02:14:59.780 I'll put all those numbers on the number line. The places where am I 02:14:59.779 --> 02:15:04.859 rational expression is equal to zero, and the place where my rational expression doesn't 02:15:04.859 --> 02:15:13.630 exist, then I can start in with test values. For example, x equals minus four, zero and 02:15:13.630 --> 02:15:20.859 to work. If I plug those values into this expression here, I get a negative answer a 02:15:20.859 --> 02:15:26.109 negative answer and a positive answer. The reason I need to conclude the values on my 02:15:26.109 --> 02:15:32.009 number line where my denominators zero is because I can my expression can switch from 02:15:32.010 --> 02:15:36.630 negative to positive by passing through a place where my rational expression doesn't 02:15:36.630 --> 02:15:41.859 exist, as well as passing by passing flew through a place where my rational expression 02:15:41.859 --> 02:15:43.219 is equal to zero. 02:15:43.220 --> 02:15:49.570 Now I'm looking for where my original expression was less than or equal to zero. So that means 02:15:49.569 --> 02:15:56.109 I want the places on the number line where my expression is equal to zero, and also the 02:15:56.109 --> 02:15:59.589 places where it's negative. 02:15:59.590 --> 02:16:06.090 So my final answer is x is less than one, or an interval notation, negative infinity 02:16:06.090 --> 02:16:08.550 to one. 02:16:08.550 --> 02:16:14.630 In this video, we solved polynomial and rational inequalities by making a number line and you 02:16:14.630 --> 02:16:19.869 think test values to make a sine chart. 02:16:19.869 --> 02:16:26.109 The distance formula can be used to find the distance between two points, if we know their 02:16:26.109 --> 02:16:27.109 coordinates, 02:16:27.109 --> 02:16:34.039 if this first point has coordinates, x one, y one, and the second point has coordinates 02:16:34.040 --> 02:16:42.010 x two, y two, then the distance between them is given by the formula, the square root of 02:16:42.010 --> 02:16:49.200 x two minus x one squared plus y two minus y one squared. 02:16:49.200 --> 02:16:55.739 This formula actually comes from the Pythagorean Theorem, let me draw a right triangle with 02:16:55.738 --> 02:17:01.399 these two points as two of its vertices, 02:17:01.399 --> 02:17:09.148 then the length of this side is the difference between the 2x coordinates. Similarly, the 02:17:09.148 --> 02:17:14.629 length of this vertical side is the difference between the two y coordinates. 02:17:14.629 --> 02:17:20.170 Now that Pythagoras theorem says that for any right triangle, with sides labeled A and 02:17:20.170 --> 02:17:27.019 B, and hypotony is labeled C, we have that a squared plus b squared equals c squared. 02:17:27.019 --> 02:17:31.728 Well, if we apply that to this triangle here, 02:17:31.728 --> 02:17:38.010 this hypotony News is the distance between the two points that we're looking for. 02:17:38.010 --> 02:17:45.818 So but Tyrion theorem says, the square of this side length, that is x two minus x one 02:17:45.818 --> 02:17:55.149 squared, plus the square of this side length, which is y two minus y one squared has to 02:17:55.149 --> 02:18:01.658 equal the square of the hypothesis, that is d squared. 02:18:01.658 --> 02:18:07.148 taking the square root of both sides of this equation, we get the square root of x two 02:18:07.148 --> 02:18:13.769 minus x one squared plus y two minus y one squared is equal to d, 02:18:13.769 --> 02:18:18.439 we don't have to worry about using plus or minus signs when we take the square root because 02:18:18.439 --> 02:18:21.260 distance is always positive. 02:18:21.260 --> 02:18:26.639 So we've derived our distance formula. Now let's use it as an example. 02:18:26.638 --> 02:18:34.589 Let's find the distance between the two points, negative one five, for negative one, five, 02:18:34.590 --> 02:18:39.079 maybe over here, and for two, 02:18:39.079 --> 02:18:48.359 this notation just means that P is the point with coordinates negative one five, and q 02:18:48.359 --> 02:18:52.699 is the point with coordinates for two. 02:18:52.699 --> 02:18:54.590 We have the distance formula, 02:18:54.590 --> 02:19:02.299 let's think of P being the point with coordinates x one y one, and Q being the point with coordinates 02:19:02.299 --> 02:19:04.278 x two y two. 02:19:04.279 --> 02:19:10.179 But as we'll see, it really doesn't matter which one is which, plugging into our formula, 02:19:10.179 --> 02:19:23.609 we get d is the square root of n x two minus x one. So that's four minus negative one squared, 02:19:23.609 --> 02:19:26.769 and then we add 02:19:26.769 --> 02:19:37.349 y two minus y one, so that's two minus five squared. 02:19:37.349 --> 02:19:41.710 Working out the arithmetic a little bit for minus negative one, that's four plus one or 02:19:41.709 --> 02:19:49.139 five squared, plus negative three squared. So that's the square root of 25 plus nine 02:19:49.139 --> 02:19:55.659 are the square root of 34. Let's see what would have happened if we'd call this first 02:19:55.659 --> 02:20:00.200 point, x two y two instead, and a second. 02:20:00.200 --> 02:20:06.390 Why'd x one y one, then we would have gotten the same distance formula, but we would have 02:20:06.389 --> 02:20:12.180 taken negative one minus four 02:20:12.180 --> 02:20:22.130 and added the difference of y's squared. So that's why two five minus two squared. 02:20:22.129 --> 02:20:27.139 That gives us the square root of negative five squared plus three squared, which works 02:20:27.139 --> 02:20:31.180 out to the same thing. 02:20:31.180 --> 02:20:36.771 In this video, we use the distance formula to find the distance between two points. When 02:20:36.771 --> 02:20:40.579 writing down the distance formula, sometimes students forget whether this is a plus or 02:20:40.578 --> 02:20:45.488 a minus and whether these are pluses or minuses. One way to remember is to think that the distance 02:20:45.488 --> 02:20:51.629 formula comes from the Pythagorean Theorem. That's why there's a plus in here. 02:20:51.629 --> 02:20:56.239 And then a minus in here, because we're finding the difference between the coordinates to 02:20:56.239 --> 02:20:59.421 find the lengths of the sides. 02:20:59.421 --> 02:21:05.699 The midpoint formula helps us find the coordinates of the midpoint of a line segment, as long 02:21:05.699 --> 02:21:09.819 as we know the coordinates of the endpoints. 02:21:09.818 --> 02:21:15.328 So let's call the coordinates of this endpoint x one, y one, and the coordinates of this 02:21:15.328 --> 02:21:17.600 other endpoint x two, y two, 02:21:17.600 --> 02:21:25.750 the x coordinate of the midpoint is going to be exactly halfway in between the x coordinates 02:21:25.750 --> 02:21:31.068 of the endpoints. To get a number halfway in between two other numbers, we just take 02:21:31.068 --> 02:21:34.340 the average. 02:21:34.340 --> 02:21:42.238 Similarly, the y coordinate of this midpoint is going to be exactly halfway in between 02:21:42.238 --> 02:21:47.430 the y coordinates of the endpoints. So the y coordinate of the midpoint is going to be 02:21:47.430 --> 02:21:53.050 the average of those y coordinates. 02:21:53.049 --> 02:22:00.938 So we see that the coordinates of the midpoint are x one plus x two over two, y one plus 02:22:00.939 --> 02:22:04.059 y two over two. 02:22:04.059 --> 02:22:07.028 Let's use this midpoint formula in an example, 02:22:07.029 --> 02:22:17.069 we want to find the midpoint of the segment between the points, negative one, five, 02:22:17.068 --> 02:22:20.988 and four 02:22:20.988 --> 02:22:24.090 to 02:22:24.090 --> 02:22:29.510 me draw the line segment between them. So visually, the midpoint is going to be somewhere 02:22:29.510 --> 02:22:36.248 around here. But to find its exact coordinates, we're going to use x one plus x two over two, 02:22:36.248 --> 02:22:44.209 and y one plus y two over two, where this first point is coordinates, x one, y one, 02:22:44.209 --> 02:22:49.788 and the second point has coordinates x two y two, it doesn't actually matter which point 02:22:49.789 --> 02:22:54.600 you decide is x one y one, which one is x two, y two, the formula will still give you 02:22:54.600 --> 02:22:56.460 the same answer for the midpoint. 02:22:56.459 --> 02:23:06.349 So let's see, I take my average of my x coordinates. So that's negative one, plus four over two, 02:23:06.350 --> 02:23:15.369 and the average of my Y coordinates, so that's five plus two over two, and I get three halves, 02:23:15.369 --> 02:23:19.069 seven halves, as the coordinates of my midpoint. 02:23:19.068 --> 02:23:25.449 In this video, we use the midpoint formula to find the midpoint of a line segment, just 02:23:25.450 --> 02:23:30.351 by taking the average of the x coordinates and the average of the y coordinates. This 02:23:30.351 --> 02:23:36.149 video is about graphs and equations of circles. Suppose we want to find the equation of a 02:23:36.148 --> 02:23:42.439 circle of radius five, centered at the point three, two 02:23:42.439 --> 02:23:47.510 will look something like this. 02:23:47.510 --> 02:23:53.420 For any point, x, y on the circle, we know that 02:23:53.420 --> 02:24:00.930 the distance of that point xy from the center is equal to the radius that is five from the 02:24:00.930 --> 02:24:07.398 distance formula, that distance of five is equal to the square root of the difference 02:24:07.398 --> 02:24:11.498 of the x coordinates. That's x minus three 02:24:11.498 --> 02:24:20.199 squared plus the difference of the y coordinates, that's y minus two squared. 02:24:20.199 --> 02:24:31.239 And if I square both sides of that equation, I get five squared is this square root squared. 02:24:31.239 --> 02:24:40.238 In other words, 25 is equal to x minus three squared plus y minus two squared. Since the 02:24:40.238 --> 02:24:43.478 square root and the squared undo each other. 02:24:43.478 --> 02:24:49.519 A lot of times people will write the X minus three squared plus y minus two squared on 02:24:49.520 --> 02:24:53.899 the left side of the equation and the 25 on the right side of the equation. This is the 02:24:53.898 --> 02:24:58.099 standard form for the equation of the circle. 02:24:58.100 --> 02:24:59.998 The same reasoning can be generalized 02:24:59.998 --> 02:25:07.119 Find the general equation for a circle with radius R centered at a point HK. 02:25:07.120 --> 02:25:14.609 For any point with coordinates, x, y on the circle, the distance between the point with 02:25:14.609 --> 02:25:22.130 coordinates x, y and the point with coordinates HK is equal to the radius r. So we have that 02:25:22.129 --> 02:25:28.709 the distance is equal to r, but by the distance formula, that's the square root of the difference 02:25:28.709 --> 02:25:36.648 between the x coordinates x minus h squared plus the difference in the y coordinates y 02:25:36.648 --> 02:25:48.559 minus k squared. squaring both sides as before, we get r squared is the square root squared, 02:25:48.559 --> 02:25:53.078 canceling the square root and the squared, and rearranging the equation, that gives us 02:25:53.078 --> 02:26:01.090 x minus h squared plus y minus k squared equals r squared. That's the general formula for 02:26:01.090 --> 02:26:06.969 a circle with radius r, and center HK. 02:26:06.969 --> 02:26:13.359 Notice that the coordinates h and k are subtracted here, the two squares are added because they 02:26:13.360 --> 02:26:18.760 are in the distance formula, and the radius is squared on the other side, if you remember 02:26:18.760 --> 02:26:23.960 this general formula, that makes it easy to write down the equation for a circle, for 02:26:23.959 --> 02:26:29.238 example, if we want the equation for a circle, 02:26:29.238 --> 02:26:35.618 of radius six, and center at zero, negative three, 02:26:35.619 --> 02:26:45.670 then we have our equal six, and h k is our zero negative three. So plugging into the 02:26:45.670 --> 02:26:47.040 formula, 02:26:47.040 --> 02:26:57.130 we get x minus zero squared, plus y minus negative three squared equals six squared, 02:26:57.129 --> 02:27:04.299 or simplified this is x squared plus y plus three squared equals 36. 02:27:04.299 --> 02:27:09.748 Suppose we're given an equation like this one, and we want to decide if it's the equation 02:27:09.748 --> 02:27:13.760 of a circle, and if so what's the center was the radius. 02:27:13.760 --> 02:27:21.728 Well, this equation matches the form for a circle, x minus h squared plus y minus k squared 02:27:21.728 --> 02:27:30.129 equals r squared. If we let h, b five, Why be negative sex since that way, subtracting 02:27:30.129 --> 02:27:34.349 a negative six is like adding a six, 02:27:34.350 --> 02:27:41.140 five must be our r squared. So that means that R is the square root of five. So this 02:27:41.139 --> 02:27:49.500 is our radius. And our center is the point five, negative six. And this is indeed the 02:27:49.500 --> 02:27:52.109 equation for a circle, 02:27:52.109 --> 02:27:57.930 which we could then graph by putting down the center and estimating the radius, which 02:27:57.930 --> 02:28:03.000 is a little bit more than two. 02:28:03.000 --> 02:28:08.859 This equation might not look like the equation of a circle, but it actually can be transformed. 02:28:08.859 --> 02:28:13.498 To look like the equation of a circle, we're trying to make it look like something of the 02:28:13.498 --> 02:28:19.270 form x minus h squared plus y minus k squared equals r squared. 02:28:19.270 --> 02:28:24.020 First, I'd like to get rid of the coefficients in front of the x squared and the y squared, 02:28:24.020 --> 02:28:30.488 so I'm going to divide everything by nine. This gives me x squared plus y squared plus 02:28:30.488 --> 02:28:38.609 8x minus two y plus four equals zero. Next, I'm going to group the x terms together the 02:28:38.609 --> 02:28:45.238 x squared and the 8x. So write x squared plus 8x. I'll group the y terms together the Y 02:28:45.238 --> 02:28:49.119 squared minus two y. 02:28:49.120 --> 02:28:54.750 And I'll subtract the four over to the other side. 02:28:54.750 --> 02:28:59.068 This still doesn't look much like the equation of a circle. But as the next step, I'm going 02:28:59.068 --> 02:29:05.288 to do something called completing the square, I'm going to take this coefficient of x eight 02:29:05.289 --> 02:29:12.210 divided by two and then square it. In other words, eight over two squared, that's four 02:29:12.209 --> 02:29:18.608 squared, which is 16. I'm going to add 16 to both sides of my equation. 02:29:18.609 --> 02:29:22.318 So the 16 here, 02:29:22.318 --> 02:29:24.439 and on the other side, 02:29:24.439 --> 02:29:31.488 now I'm going to do the same thing to the coefficient of y, negative two divided by 02:29:31.488 --> 02:29:38.260 two is negative one, square that and I get one. So now I'm going to add a one to both 02:29:38.260 --> 02:29:44.068 sides, but I'll put it near the y terms. So add a one there. And then to keep things balanced, 02:29:44.068 --> 02:29:50.639 I have to add it to the other side. On the right side, I have the number 13. On the left 02:29:50.639 --> 02:29:59.439 side, I can wrap up this expression x squared plus 8x plus 16 into x plus four squared. 02:29:59.439 --> 02:30:07.460 To convince you, that's correct. If we multiply out x plus four times x plus four, we get 02:30:07.459 --> 02:30:16.919 x squared plus 4x plus 4x plus 16, or x squared plus 8x plus 16, the same thing we have right 02:30:16.920 --> 02:30:25.260 here. Similarly, we can wrap up y squared minus two y plus one as 02:30:25.260 --> 02:30:32.340 y minus one squared. Again, I'll work out the distribution down here, that's y squared 02:30:32.340 --> 02:30:40.030 minus y minus y plus one, or y squared minus two y plus one, just like we have up here. 02:30:40.030 --> 02:30:46.989 If you're wondering how I knew to use four and minus one, the four came from taking half 02:30:46.988 --> 02:30:53.988 of eight, and the minus one came from taking half the negative two. Now we have an equation 02:30:53.988 --> 02:31:01.629 for a circle and standard form. And we can read off the center, which is negative four, 02:31:01.629 --> 02:31:10.429 negative one, and the radius, which is the square root of 13. 02:31:10.430 --> 02:31:14.979 It might seem like magic that this trick of taking half of this coefficient and squaring 02:31:14.978 --> 02:31:20.568 it and adding it to both sides lets us wrap up this expression into this expression a 02:31:20.568 --> 02:31:26.988 perfect square. But to see why that works, let's take a look at what happens if we expand 02:31:26.988 --> 02:31:33.760 out x minus h squared, the thing that we're trying to get to. So if we expand that out, 02:31:33.760 --> 02:31:43.609 x minus h squared, which is x minus h times x minus h is, which multiplies out to x squared 02:31:43.609 --> 02:31:53.430 minus HX minus h x plus h squared, or x squared minus two h x plus h squared. Now if we start 02:31:53.430 --> 02:32:01.090 out with this part, with some x squared term and some coefficient of x, we're trying to 02:32:01.090 --> 02:32:06.409 decide what to add on in order to wrap it up into x minus h squared. The thing to add 02:32:06.409 --> 02:32:13.430 on is h squared here, which comes from half of this coefficient squared, right, because 02:32:13.430 --> 02:32:18.689 half of negative two H is a negative H, square that you get the H squared. And then when 02:32:18.689 --> 02:32:26.738 we do wrap it up, it's half of this coefficient of x, that appears right here. 02:32:26.738 --> 02:32:30.988 This trick of completing the square is really handy for turning an equation for a circle 02:32:30.988 --> 02:32:37.260 in disguise, into the standard equation for the circle. In this video, we found the standard 02:32:37.260 --> 02:32:44.728 equation for a circle x minus h squared plus y minus k squared equals r squared, where 02:32:44.728 --> 02:32:52.159 r is the radius, and h k is the center. 02:32:52.159 --> 02:32:56.648 We also showed a method of completing the square. 02:32:56.648 --> 02:33:01.840 When you have an equation for a circle in disguise, completing the square will help 02:33:01.840 --> 02:33:06.279 you rewrite it into the standard form. 02:33:06.279 --> 02:33:10.979 This video is about graphs and equations of lines. 02:33:10.978 --> 02:33:16.799 Here we're given the graph of a line, we want to find the equation, one standard format 02:33:16.799 --> 02:33:25.059 for the equation of a line is y equals mx plus b. here am represents the slope, and 02:33:25.059 --> 02:33:34.549 B represents the y intercept, the y value where the line crosses the y axis, the slope 02:33:34.549 --> 02:33:41.228 is equal to the rise over the run. Or sometimes this is written as the change in y values 02:33:41.228 --> 02:33:49.549 over the change in x values. Or in other words, y two minus y one over x two minus x one, 02:33:49.549 --> 02:33:56.509 where x one y one and x two y two are points on the line. 02:33:56.510 --> 02:34:01.600 While we could use any two points on the line, to find the slope, it's convenient to use 02:34:01.600 --> 02:34:08.590 points where the x and y coordinates are integers, that is points where the line passes through 02:34:08.590 --> 02:34:13.630 grid points. So here would be one convenient point to use. And here's another convenient 02:34:13.629 --> 02:34:15.639 point to use. 02:34:15.639 --> 02:34:22.068 The coordinates of the first point are one, two, and the next point, this is let's say, 02:34:22.068 --> 02:34:29.319 five, negative one. Now I can find the slope by looking at the rise over the run. So as 02:34:29.319 --> 02:34:35.629 I go through a run of this distance, I go through a rise of that distance especially 02:34:35.629 --> 02:34:41.078 gonna be a negative rise or a fall because my line is pointing down. So let's see counting 02:34:41.078 --> 02:34:48.219 off squares. This is a run of 1234 squares and a rise of 123. So negative three, so my 02:34:48.219 --> 02:34:52.658 slope is going to be 02:34:52.658 --> 02:34:56.270 negative three over four. 02:34:56.270 --> 02:35:00.510 I got that answer by counting squares, but I could have also gotten it by 02:35:00.510 --> 02:35:05.630 Looking at the difference in my y values over the difference in my x values, that is, I 02:35:05.629 --> 02:35:08.179 could have done 02:35:08.180 --> 02:35:15.898 negative one minus two, that's from my difference in Y values, and divide that by my difference 02:35:15.898 --> 02:35:18.459 in x values, 02:35:18.459 --> 02:35:26.188 which is five minus one, that gives me negative three over four, as before. 02:35:26.189 --> 02:35:30.750 So my M is negative three fourths. 02:35:30.750 --> 02:35:36.219 Now I need to figure out the value of b, my y intercept, well, I could just read it off 02:35:36.219 --> 02:35:41.868 the graph, it looks like approximately 2.75. But if I want to be more accurate, I can again 02:35:41.869 --> 02:35:47.539 use a point that has integer coordinates that I know it's exact coordinates. So either this 02:35:47.539 --> 02:35:53.689 point or that point, let's try this point. And I can start off with my equation y equals 02:35:53.689 --> 02:36:02.389 mx plus b, that is y equals negative three fourths x plus b. And I can plug in the point 02:36:02.389 --> 02:36:12.118 one, two, for my x and y. So that gives me two equals negative three fourths times one 02:36:12.119 --> 02:36:20.078 plus b, solving for B. Let's see that's two equals negative three fourths plus b. So add 02:36:20.078 --> 02:36:28.090 three fourths to both sides, that's two plus three fourths equals b. So b is eight fourths 02:36:28.090 --> 02:36:33.500 plus three fourths, which is 11. Four switches actually, just what I eyeballed it today. 02:36:33.500 --> 02:36:40.559 So now I can write out my final equation for my line y equals negative three fourths x 02:36:40.559 --> 02:36:48.289 plus 11 fourths by plugging in for m and b. Next, let's find the equation for this horizontal 02:36:48.290 --> 02:36:49.420 line. 02:36:49.420 --> 02:36:57.158 a horizontal line has slope zero. So if we think of it as y equals mx plus b, m is going 02:36:57.158 --> 02:37:03.898 to be zero. In other words, it's just y equals b, y is some constant. So if we can figure 02:37:03.898 --> 02:37:08.139 out what that that constant y value is, it looks like it's 02:37:08.139 --> 02:37:14.818 two, let's see this three, three and a half, we can just write down the equation directly, 02:37:14.818 --> 02:37:18.998 y equals 3.5. 02:37:18.998 --> 02:37:24.898 For a vertical line, like this one, it doesn't really have a slope. I mean, if you tried 02:37:24.898 --> 02:37:26.449 to do the rise over the run, 02:37:26.450 --> 02:37:32.340 there's no run. So you'd I guess you'd be dividing by zero and get an infinite slope. 02:37:32.340 --> 02:37:39.859 But But instead, we just think of it as an equation of the form x equals something. And 02:37:39.859 --> 02:37:45.199 in this case, x equals negative two, notice that all of the points on our line have the 02:37:45.199 --> 02:37:50.380 same x coordinate of negative two and the y coordinate can be anything. 02:37:50.379 --> 02:37:54.879 So this is how we write the equation for a vertical line. 02:37:54.879 --> 02:37:59.299 In this example, we're not shown a graph of the line, we're just get told that it goes 02:37:59.299 --> 02:38:04.629 through two points. But knowing that I go through two points is enough to find the equation 02:38:04.629 --> 02:38:12.608 for the line. First, we can find the slope by taking the difference in Y values over 02:38:12.609 --> 02:38:20.090 the difference in x values. So that's negative three minus two over four minus one, which 02:38:20.090 --> 02:38:24.319 is negative five thirds. 02:38:24.319 --> 02:38:26.329 So we can use the 02:38:26.329 --> 02:38:34.648 standard equation for the line, this is called the slope intercept form. 02:38:34.648 --> 02:38:42.719 And we can plug in negative five thirds. And we can use one point, either one will do will 02:38:42.719 --> 02:38:48.488 still get the same final answer. So let's use one two and plug that in to get two equals 02:38:48.488 --> 02:38:57.779 negative five thirds times one plus b. And so B is two plus five thirds, which is six 02:38:57.779 --> 02:39:03.550 thirds plus five thirds, which is 11 thirds. So our equation is y equals negative five 02:39:03.549 --> 02:39:07.389 thirds x plus 11 thirds. 02:39:07.389 --> 02:39:11.728 This is method one. 02:39:11.728 --> 02:39:17.119 method two uses a slightly different form of the equation, it's called the point slope 02:39:17.120 --> 02:39:25.680 form and it goes y minus y naught is equal to m times x minus x naught, where x naught 02:39:25.680 --> 02:39:33.380 y naught is a point on the line and again is the slope. So we calculate the slope the 02:39:33.379 --> 02:39:40.309 same way by taking a difference in Y values over a difference in x values. But then, we 02:39:40.309 --> 02:39:43.698 can simply plug in any point. 02:39:43.699 --> 02:39:53.039 For example, the point one two will work we can plug one in for x naught and two in for 02:39:53.039 --> 02:39:59.420 Y not in this point slope form. That gives us y 02:39:59.420 --> 02:40:00.420 minus 02:40:00.420 --> 02:40:06.719 Two is equal to minus five thirds x minus one. 02:40:06.719 --> 02:40:11.090 Notice that these two equations, while they may look different, are actually equivalent, 02:40:11.090 --> 02:40:25.380 because if I distribute the negative five thirds, and then add the two to both sides, 02:40:25.379 --> 02:40:28.529 I get the same equation as above. 02:40:28.530 --> 02:40:37.100 So we've seen two ways of finding the equation for the line using the slope intercept form. 02:40:37.100 --> 02:40:40.199 And using the point slope form. 02:40:40.199 --> 02:40:48.920 In this video, we saw that you can find the equation for a line, if you know the slope. 02:40:48.920 --> 02:40:52.158 And you know one point, 02:40:52.158 --> 02:40:57.760 you can also find the equation for the line if you know two points, because you can use 02:40:57.760 --> 02:41:03.529 the two points to get the slope and then plug in one of those points. To figure out the 02:41:03.529 --> 02:41:06.430 rest of the equation. 02:41:06.430 --> 02:41:08.898 We saw two standard forms for the equation of a line 02:41:08.898 --> 02:41:19.738 the slope intercept form y equals mx plus b, where m is the slope, and B is the y intercept. 02:41:19.738 --> 02:41:27.398 And the point slope form y minus y naught equals m times x minus x naught, where m again 02:41:27.398 --> 02:41:35.328 is the slope and x naught y naught is a point on the line. 02:41:35.328 --> 02:41:43.219 This video is about finding parallel and perpendicular lines. Suppose we have a line of slope three 02:41:43.219 --> 02:41:47.469 fourths, in other words, the rise 02:41:47.469 --> 02:41:50.590 over the run 02:41:50.590 --> 02:41:52.529 is three over four, 02:41:52.529 --> 02:42:01.119 than any other line that's parallel to this line will have the same slope, the same fries 02:42:01.119 --> 02:42:03.939 over run. 02:42:03.939 --> 02:42:09.909 So that's our first fact to keep in mind, parallel lines have the same slopes. Suppose 02:42:09.909 --> 02:42:15.000 on the other hand, we want to find a line perpendicular to our original line with its 02:42:15.000 --> 02:42:18.129 original slope of three fourths. 02:42:18.129 --> 02:42:25.358 A perpendicular line, in other words, align at a 90 degree angle to our original line 02:42:25.359 --> 02:42:32.039 will have a slope, that's the negative reciprocal of the opposite reciprocal of our original 02:42:32.039 --> 02:42:37.579 slope. So we take the reciprocal of three for us, that's four thirds, and we make it 02:42:37.578 --> 02:42:41.359 its opposite by changing it from positive to negative. 02:42:41.359 --> 02:42:49.738 So I'll write this as a principle that perpendicular lines have opposite reciprocal slopes, to 02:42:49.738 --> 02:42:54.719 get the hang of what it means to be an opposite reciprocal, let's look at a few examples. 02:42:54.719 --> 02:43:01.288 So here's the original slope, and this will be the opposite reciprocal, which would represent 02:43:01.289 --> 02:43:07.610 the slope of our perpendicular line. So for example, if one was to the opposite reciprocal, 02:43:07.610 --> 02:43:14.319 reciprocal of two is one half opposite means I change the sign, if the first slope turned 02:43:14.318 --> 02:43:23.260 out to be, say, negative 1/3, the reciprocal of 1/3 is three over one or just three. And 02:43:23.260 --> 02:43:29.359 the opposite means I change it from a negative to a positive, so my perpendicular slope would 02:43:29.359 --> 02:43:32.189 be would be three. 02:43:32.189 --> 02:43:38.609 One more example, if I started off with a slope of, say, seven halves, then the reciprocal 02:43:38.609 --> 02:43:43.890 of that would be two sevenths. And I change it to an opposite reciprocal, by changing 02:43:43.889 --> 02:43:47.590 the positive to a negative. 02:43:47.590 --> 02:43:52.469 Let's use these two principles in some examples. 02:43:52.469 --> 02:43:56.488 In our first example, we need to find the equation of a line that's parallel to this 02:43:56.488 --> 02:44:02.908 slump, and go through the point negative three two. Well, in order to get started, I need 02:44:02.908 --> 02:44:07.538 to figure out the slope of this line. So let me put this line into a more standard form 02:44:07.539 --> 02:44:14.459 the the slope intercept form. So I'll start with three y minus 4x plus six equals zero, 02:44:14.459 --> 02:44:19.829 I'm going to solve for y to put it in this this more standard form. So I'll say three 02:44:19.829 --> 02:44:29.439 y equals 4x minus six and then divide by three to get y equals four thirds x minus six thirds 02:44:29.439 --> 02:44:34.850 is divided all my turns by three, I can simplify that a little bit y equals four thirds x minus 02:44:34.850 --> 02:44:43.439 two. Now I can read off the slope of my original line, and one is four thirds. That means my 02:44:43.439 --> 02:44:51.279 slope of my new line, my parallel line will also be four thirds. So my new line will have 02:44:51.279 --> 02:44:58.060 the equation y equals four thirds ax plus b for some B, of course, B will probably not 02:44:58.059 --> 02:45:00.029 be negative two like it was for the 02:45:00.030 --> 02:45:04.189 first line, it'll be something else determined by the fact that it goes through this point 02:45:04.189 --> 02:45:10.359 negative three to, to figure out what b is, I plug in the point negative three to four, 02:45:10.359 --> 02:45:16.408 negative three for x, and two for y. So that gives me two equals four thirds times negative 02:45:16.408 --> 02:45:25.449 three plus b, and I'll solve for b. So let's see. First, let me just simplify. So two equals, 02:45:25.449 --> 02:45:32.440 this is negative 12 thirds plus b, or in other words, two is negative four plus b. So that 02:45:32.440 --> 02:45:38.479 means that B is going to be six, and by my parallel line will have the equation y equals 02:45:38.478 --> 02:45:42.340 four thirds x plus six. 02:45:42.340 --> 02:45:49.398 Next, let's find the equation of a line that's perpendicular to a given line through and 02:45:49.398 --> 02:45:54.019 go through a given point. Again, in order to get started, I need to find what the slope 02:45:54.020 --> 02:45:59.869 of this given line is. So I'll rewrite it, well, first, I'll just copy it over 6x plus 02:45:59.869 --> 02:46:06.670 three, y equals four. And then I can put it in the standard slope intercept form by solving 02:46:06.670 --> 02:46:14.879 for y. So three, y is negative 6x plus four, divide everything by three, I get y equals 02:46:14.879 --> 02:46:22.248 negative six thirds x plus four thirds. So y is negative 2x, plus four thirds. So my 02:46:22.248 --> 02:46:28.068 original slope of my original line is negative two, which means my perpendicular slope is 02:46:28.068 --> 02:46:34.028 the opposite reciprocal, so I take the reciprocal of of negative two, that's negative one half 02:46:34.029 --> 02:46:39.380 and I change the sign so that gives me one half as the slope of my perpendicular line. 02:46:39.379 --> 02:46:47.209 Now, my new line, I know is going to be y equals one half x plus b for some B, and I 02:46:47.209 --> 02:46:57.148 can plug in my point on my new line, so one for y and four for x, and solve for b, so 02:46:57.148 --> 02:47:04.778 I get one equals two plus b, So b is negative one. And so my new line has equation one half 02:47:04.779 --> 02:47:07.520 x minus one. 02:47:07.520 --> 02:47:14.439 These next two examples are a little bit different, because now we're looking at a line parallel 02:47:14.439 --> 02:47:23.359 to a completely horizontal line, let me draw this line y equals three, so the y coordinate 02:47:23.359 --> 02:47:29.270 is always three, which means that my line will be a horizontal line through that height 02:47:29.270 --> 02:47:36.329 at height y equals three, if I want something parallel to this line, it will also be a horizontal 02:47:36.329 --> 02:47:42.799 line. Since it goes through the point negative two, one, C, negative two, one has to go through 02:47:42.799 --> 02:47:45.590 that point there, 02:47:45.590 --> 02:47:51.309 it's gonna have always have a y coordinate of one the same as the y coordinate of the 02:47:51.309 --> 02:47:55.510 point it goes through, so my answer will just be y equals one. 02:47:55.510 --> 02:48:06.260 In the next example, we want a line that's perpendicular to the line y equals four another 02:48:06.260 --> 02:48:12.328 horizontal line y equals four, but perpendicular means I'm going to need a vertical line. So 02:48:12.328 --> 02:48:16.818 I need a vertical line that goes through the point three, four. 02:48:16.818 --> 02:48:27.748 Okay, and so I'm going to draw a vertical line there. Now vertical lines have the form 02:48:27.748 --> 02:48:32.949 x equals something for the equation. And to find what x equals, I just need to look at 02:48:32.950 --> 02:48:38.359 the x coordinate of the point I'm going through this point three four, so that x coordinate 02:48:38.359 --> 02:48:45.059 is three and all the points on this, this perpendicular and vertical line have X coordinate 02:48:45.059 --> 02:48:53.748 three so my answer is x equals three. 02:48:53.748 --> 02:48:58.148 In this video will use the fact that parallel lines have the same slope and perpendicular 02:48:58.148 --> 02:49:03.250 lines have opposite reciprocal slopes, to find the equations of parallel and perpendicular 02:49:03.250 --> 02:49:05.510 lines. 02:49:05.510 --> 02:49:09.988 This video introduces functions and their domains and ranges. 02:49:09.988 --> 02:49:17.930 A function is a correspondence between input numbers usually the x values and output numbers, 02:49:17.930 --> 02:49:25.079 usually the y values, that sends each input number to exactly one output number. Sometimes 02:49:25.079 --> 02:49:28.260 a function is thought of as a rule or machine 02:49:28.260 --> 02:49:38.880 in which you can feed in x values as input and get out y values as output. So, non mathematical 02:49:38.879 --> 02:49:44.658 example of a function might be the biological mother function, which takes as input any 02:49:44.658 --> 02:49:49.959 person and it gives us output their biological mother. 02:49:49.959 --> 02:49:57.349 This function satisfies the condition that each input number object person in this case, 02:49:57.350 --> 02:50:01.068 get sent to exactly one output per 02:50:01.068 --> 02:50:07.059 Because if you take any person, they just have one biological mother. So that rule does 02:50:07.059 --> 02:50:14.049 give you a function. But if I change things around and just use the the mother function, 02:50:14.049 --> 02:50:19.139 which sends to each person, their mother, that's no longer going to be a function because 02:50:19.139 --> 02:50:23.250 there are some people who have more than one mother, right, they could have a biological 02:50:23.250 --> 02:50:28.318 mother and adopted mother, or a mother and a stepmother, or any number of situations. 02:50:28.318 --> 02:50:33.590 So since there's, there's at least some people who you would put it as input, and then you'd 02:50:33.590 --> 02:50:39.029 get like more than one possible output that violates this, this rule of functions, that 02:50:39.029 --> 02:50:43.810 would not be a function. Now, most of the time, we'll work with functions that are described 02:50:43.809 --> 02:50:48.948 with equations, not in terms of mothers. So for example, we can have the function y equals 02:50:48.949 --> 02:50:51.329 x squared plus one. 02:50:51.329 --> 02:50:57.409 This can also be written as f of x equals x squared plus one. Here, f of x is function 02:50:57.409 --> 02:51:02.510 notation that stands for the output value of y. 02:51:02.510 --> 02:51:07.760 Notice that this notation is not representing multiplication, we're not multiplying f by 02:51:07.760 --> 02:51:14.309 x, instead, we're going to be putting in a value for x as input and getting out a value 02:51:14.309 --> 02:51:21.670 of f of x or y. For example, if we want to evaluate f of two, we're plugging in two as 02:51:21.670 --> 02:51:30.248 input for x either in this equation, or in that equation. Since F of two means two squared 02:51:30.248 --> 02:51:38.629 plus one, f of two is going to equal five. Similarly, f of five means I plug in five 02:51:38.629 --> 02:51:46.719 for x, so that's gonna be five squared plus one, or 26. Sometimes it's useful to evaluate 02:51:46.719 --> 02:51:52.420 a function on a more complicated expression involving other variables. In this case, remember, 02:51:52.420 --> 02:51:57.748 the functions value on any expression, it's just what you what you get when you plug in 02:51:57.748 --> 02:52:00.680 that whole expression for x. 02:52:00.680 --> 02:52:09.389 So f of a plus three is going to be the quantity a plus three squared plus one, 02:52:09.389 --> 02:52:17.029 we could rewrite that as a squared plus six a plus nine plus one, or a squared plus six 02:52:17.030 --> 02:52:19.649 a plus 10. 02:52:19.648 --> 02:52:25.090 When evaluating a function on a complex expression, it's important to keep the parentheses when 02:52:25.090 --> 02:52:32.119 you plug in for x. That way, you evaluate the function on the whole expression. For 02:52:32.119 --> 02:52:35.209 example, it would be wrong 02:52:35.209 --> 02:52:42.590 to write f of a plus three equals a plus three squared plus one without the parentheses, 02:52:42.590 --> 02:52:47.978 because that would imply that we were just squaring the three and not the whole expression, 02:52:47.978 --> 02:52:52.379 a plus three. 02:52:52.379 --> 02:52:58.009 Sometimes a function is described with a graph instead of an equation. In this example, this 02:52:58.010 --> 02:53:04.939 graph is supposed to represent the function g of x. Not all graphs actually represent 02:53:04.939 --> 02:53:11.050 functions. For example, the graph of a circle doesn't represent a function. That's because 02:53:11.049 --> 02:53:15.948 the graph of a circle violates the vertical line test, you can draw a vertical line and 02:53:15.949 --> 02:53:18.289 intersect the graph in more than one point. 02:53:18.289 --> 02:53:24.130 But our graph, it left satisfies the vertical line test, any vertical line intersects the 02:53:24.129 --> 02:53:29.028 graph, and at most one point, that means is a function, because every x value will have 02:53:29.029 --> 02:53:37.439 at most one y value that corresponds to it. Let's evaluate g have to note that two is 02:53:37.439 --> 02:53:44.380 an x value. And we'll use the graph to find the corresponding y value. So I look for two 02:53:44.379 --> 02:53:51.278 on the x axis, and find the point on the graph that has that x value that's right here. Now 02:53:51.279 --> 02:53:58.630 I can look at the y value of that point looks like three, and therefore gf two is equal 02:53:58.629 --> 02:54:05.559 to three. If I try to do the same thing to evaluate g of five, I run into trouble. Five 02:54:05.559 --> 02:54:11.648 is an x value, I look for it on the x axis, but there's no point on the graph that has 02:54:11.648 --> 02:54:13.778 that x value. 02:54:13.779 --> 02:54:21.050 Therefore, g of five is undefined, or we can say it does not exist. The question of what 02:54:21.049 --> 02:54:27.670 x values and y values make sense for a function leads us to a discussion of domain and range. 02:54:27.670 --> 02:54:32.939 The domain of a function is all possible x values that makes sense for that function. 02:54:32.939 --> 02:54:36.898 The range is the y values that make sense for the function. 02:54:36.898 --> 02:54:42.180 In this example, we saw that the x value of five didn't have a corresponding y value for 02:54:42.180 --> 02:54:49.090 this function. So the x value of five is not in the domain of our function J. To find the 02:54:49.090 --> 02:54:54.389 x values in the domain, we have to look at the x values that correspond to points on 02:54:54.389 --> 02:55:00.228 the graph. One way to do that is to take the shadow or projection of the graph 02:55:00.228 --> 02:55:07.969 onto the x axis and see what x values are hit. It looks like we're hitting all x values 02:55:07.969 --> 02:55:15.538 starting at negative eight, and continuing up to four. So our domain 02:55:15.539 --> 02:55:21.289 is the x's between negative eight, and four, including those endpoints. Or we can write 02:55:21.289 --> 02:55:27.248 this in interval notation as negative eight comma four with square brackets. 02:55:27.248 --> 02:55:31.559 To find the range of the function, we look for the y values corresponding to points on 02:55:31.559 --> 02:55:35.278 this graph, we can do that. 02:55:35.279 --> 02:55:42.979 By taking the shadow or projection of the graph onto the y axis, 02:55:42.978 --> 02:55:51.398 we seem to be hitting our Y values from negative five, up through three. 02:55:51.398 --> 02:55:59.340 So our range is wise between negative five and three are an interval notation negative 02:55:59.340 --> 02:56:05.369 five, three with square brackets. If we meet a function that's described as an equation 02:56:05.369 --> 02:56:11.498 instead of a graph, one way to find the domain and range are to graph the function. But it's 02:56:11.498 --> 02:56:17.478 often possible to find the domain at least more quickly, by using algebraic considerations. 02:56:17.478 --> 02:56:22.328 We think about what x values that makes sense to plug into this expression, and what x values 02:56:22.328 --> 02:56:27.049 need to be excluded, because they make the algebraic expression impossible to evaluate. 02:56:27.049 --> 02:56:35.629 Specifically, to find the domain of a function, we need to exclude x values that make the 02:56:35.629 --> 02:56:38.698 denominator zero. Since we can't divide by zero, 02:56:38.699 --> 02:56:47.470 we also need to exclude x values that make an expression inside a square root sign negative. 02:56:47.469 --> 02:56:52.108 Since we can't take the square root of a negative number. In fact, we need to exclude values 02:56:52.109 --> 02:56:58.199 that make the expression inside any even root negative because we can't take an even root 02:56:58.199 --> 02:57:02.520 of a negative number, even though we can take an odd root like a cube root of a negative 02:57:02.520 --> 02:57:07.488 number. Later, when we look at logarithmic functions, we'll have some additional exclusions 02:57:07.488 --> 02:57:12.228 that we have to make. But for now, these two principles should handle all functions We'll 02:57:12.228 --> 02:57:16.639 see. So let's apply them to a couple examples. 02:57:16.639 --> 02:57:23.170 For the function in part A, we don't have any square root signs, but we do have a denominator. 02:57:23.170 --> 02:57:28.389 So we need to exclude x values that make the denominator zero. In other words, we need 02:57:28.389 --> 02:57:34.458 x squared minus 4x plus three to not be equal to zero. 02:57:34.459 --> 02:57:42.239 If we solve x squared minus 4x plus three equal to zero, we can do that by factoring. 02:57:42.238 --> 02:57:48.609 And that gives us x equals three or x equals one, so we need to exclude these values. 02:57:48.609 --> 02:57:54.359 All other x values should be fine. So if I draw the number line, I can put on one and 02:57:54.359 --> 02:58:00.689 three and just dig out a hole at both of those. And my domain includes everything else on 02:58:00.689 --> 02:58:06.828 the number line. In interval notation, this means my domain is everything from negative 02:58:06.828 --> 02:58:12.969 infinity to one, together with everything from one to three, together with everything 02:58:12.969 --> 02:58:15.809 from three to infinity. 02:58:15.809 --> 02:58:21.219 In the second example, we don't have any denominator to worry about, but we do have a square root 02:58:21.219 --> 02:58:29.688 sign. So we need to exclude any x values that make three minus 2x less than zero. In other 02:58:29.689 --> 02:58:35.760 words, we can include all x values for which three minus 2x is greater than or equal to 02:58:35.760 --> 02:58:43.279 zero. Solving that inequality gives us three is greater than equal to 2x. In other words, 02:58:43.279 --> 02:58:44.789 x 02:58:44.789 --> 02:58:48.209 is less than or equal to three halves, 02:58:48.209 --> 02:58:51.010 I can draw this on the number line, 02:58:51.010 --> 02:58:54.898 or write it in interval notation. 02:58:54.898 --> 02:59:00.719 Notice that three halves is included, and that's because three minus 2x is allowed to 02:59:00.719 --> 02:59:05.629 be zero, I can take the square root of zero, that's just zero. And that's not a problem. 02:59:05.629 --> 02:59:10.238 Finally, let's look at a more complicated function that involves both the square root 02:59:10.238 --> 02:59:14.969 and denominator. Now there are two things I need to worry about. 02:59:14.969 --> 02:59:18.099 I need the denominator 02:59:18.100 --> 02:59:24.300 to not be equal to zero, and I need the stuff inside the square root side to be greater 02:59:24.299 --> 02:59:30.309 than or equal to zero. from our earlier work, we know the first condition means that x is 02:59:30.309 --> 02:59:36.488 not equal to three, and x is not equal to one. And the second condition means that x 02:59:36.488 --> 02:59:41.879 is less than or equal to three halves. Let's try both of those conditions on the number 02:59:41.879 --> 02:59:44.179 line. 02:59:44.180 --> 02:59:51.039 x is not equal to three and x is not equal to one means we've got everything except those 02:59:51.039 --> 02:59:56.680 two dug out points. And the other condition x is less than or equal to three halves means 02:59:56.680 --> 03:00:00.209 we can have three halves and everything 03:00:00.209 --> 03:00:05.060 To the left of it. Now to be in our domain and to be legit for our function, we need 03:00:05.059 --> 03:00:09.219 both of these conditions to be true. So I'm going to connect these conditions with N and 03:00:09.219 --> 03:00:13.708 N, that means we're looking for a numbers on the number line that are colored both red 03:00:13.709 --> 03:00:19.939 and blue. So I'll draw that above in purple. So that's everything from three halves to 03:00:19.939 --> 03:00:24.709 one, I have to dig out one because one was a problem for the denominator. And then I 03:00:24.709 --> 03:00:30.300 can continue for all the things that are colored both both colors red and blue. So my final 03:00:30.299 --> 03:00:37.679 domain is going to be, let's see negative infinity, up to but not including one together 03:00:37.680 --> 03:00:44.540 with one, but not including it to three halves, and I include three half since that was colored, 03:00:44.540 --> 03:00:46.199 both red and blue. Also, 03:00:46.199 --> 03:00:51.340 in this video, we talked about functions, how to evaluate them, and how to find domains 03:00:51.340 --> 03:00:57.130 and ranges. This video gives the graphs of some commonly used functions that I call the 03:00:57.129 --> 03:01:03.849 toolkit functions. The first function is the function y equals x, let's plot a few points 03:01:03.850 --> 03:01:13.248 on the graph of this function. If x is zero, y zero, if x is one, y is one, and so on y 03:01:13.248 --> 03:01:19.059 is always equal to x doesn't have to just be an integer, it can be any real number, 03:01:19.059 --> 03:01:24.398 and we'll connecting the dots, we get a straight line through the origin. 03:01:24.398 --> 03:01:30.090 Let's look at the graph of y equals x squared. If x is zero, y is zero. So we go through 03:01:30.090 --> 03:01:40.398 the origin again, if x is one, y is one, and x is negative one, y is also one, 03:01:40.398 --> 03:01:45.248 the x value of two gives a y value of four and the x value of negative two gives a y 03:01:45.248 --> 03:01:51.059 value of four also connecting the dots, we get a parabola. 03:01:51.059 --> 03:01:56.828 That is this, this function is an even function. 03:01:56.828 --> 03:02:02.840 That means it has mirror symmetry across the y axis, the left side it looks like exactly 03:02:02.840 --> 03:02:08.139 like the mirror image of the right side. That happens because when you square a positive 03:02:08.139 --> 03:02:14.760 number, like two, you get the exact same y value as when you square it's a mirror image 03:02:14.760 --> 03:02:17.699 x value of negative two. 03:02:17.699 --> 03:02:24.119 The next function y equals x cubed. I'll call that a cubic. Let's plot a few points when 03:02:24.119 --> 03:02:31.370 x is zero, y is zero. When x is one, y is one, when x is negative one, y is negative 03:02:31.370 --> 03:02:37.319 one, two goes with the point eight way up here and an x value of negative two is going 03:02:37.318 --> 03:02:42.799 to give us negative eight. If I connect the dots, I get something that looks kind of like 03:02:42.799 --> 03:02:45.049 this. 03:02:45.049 --> 03:02:53.728 This function is what's called an odd function, because it has 180 degree rotational symmetry 03:02:53.728 --> 03:02:58.688 occur around the origin. If I rotate this whole graph by 180 degrees, or in other words, 03:02:58.689 --> 03:03:04.908 turn the paper upside down, I'll get exactly the same shape. The reason this function has 03:03:04.908 --> 03:03:12.199 this odd symmetry is because when I cube a positive number, and get its y value, I get 03:03:12.199 --> 03:03:17.670 n cube the corresponding negative number to get its y value, the negative number cubed 03:03:17.670 --> 03:03:25.109 gives us exactly the negative of the the y value we get with cubing the positive number. 03:03:25.109 --> 03:03:29.828 Let's look at the next example. Y equals the square root of x. 03:03:29.828 --> 03:03:37.010 Notice that the domain of this function is just x values bigger than or equal to zero 03:03:37.010 --> 03:03:42.478 because we can't take the square root of a negative number. Let's plug in a few x values. 03:03:42.478 --> 03:03:49.559 X is zero gives y is 0x is one square root of one is one, square root of four is two, 03:03:49.559 --> 03:03:55.448 and connecting the dots, I get a function that looks like this. 03:03:55.449 --> 03:04:01.810 The absolute value function is next. Again, if we plug in a few points, x is zero goes 03:04:01.809 --> 03:04:11.079 with y equals 0x is one gives us one, the absolute value of negative one is 122 is on 03:04:11.079 --> 03:04:17.930 the graph, and the absolute value of negative two is to I'm ending up getting a V shaped 03:04:17.930 --> 03:04:24.470 graph. It also has even or a mirror symmetry. 03:04:24.469 --> 03:04:30.379 Y equals two to the x is what's known as an exponential function. That's because the variable 03:04:30.379 --> 03:04:35.648 x is in the exponent. If I plot a few points, 03:04:35.648 --> 03:04:41.930 two to the zero is one, 03:04:41.930 --> 03:04:49.709 two to the one is two, two squared is four, two to the minus one is one half. I'll plot 03:04:49.709 --> 03:04:55.850 these on my graph, you know, let me fill in a few more points. So let's see. two cubed 03:04:55.850 --> 03:04:59.908 is eight. That's way up here and negative two gives 03:04:59.908 --> 03:05:07.318 Maybe 1/4 1/8 connecting the dots, I get something that's shaped like this. 03:05:07.318 --> 03:05:11.988 You might have heard the expression exponential growth, when talking about, for example, population 03:05:11.988 --> 03:05:19.359 growth, this is function is represents exponential growth, it grows very, very steeply. Every 03:05:19.359 --> 03:05:26.689 time we increase the x coordinate by one, we double the y coordinate. 03:05:26.689 --> 03:05:32.970 We could also look at a function like y equals three dx, or sometimes y equals e to the x, 03:05:32.969 --> 03:05:40.129 where E is just a number about 2.7. These functions look very similar. It's just the 03:05:40.129 --> 03:05:45.948 bigger bass makes us rise a little more steeply. 03:05:45.949 --> 03:05:52.211 Now let's look at the function y equals one over x. It's not defined when x is zero, but 03:05:52.210 --> 03:05:58.228 I can plug in x equals one half, one over one half is to 03:05:58.228 --> 03:06:06.528 whenever one is one, and one or two is one half. By connect the dots, I get this shape 03:06:06.529 --> 03:06:11.090 in the first quadrant, but I haven't looked at negative values of x, when x is negative, 03:06:11.090 --> 03:06:18.639 whenever negative one is negative one, whenever negative half is negative two, and I get a 03:06:18.639 --> 03:06:22.318 similar looking piece in the third quadrant. 03:06:22.318 --> 03:06:27.639 This is an example of a hyperbola. 03:06:27.639 --> 03:06:34.788 And it's also an odd function, because it has that 180 degree rotational symmetry, if 03:06:34.789 --> 03:06:40.369 I turn the page upside down, it'll look exactly the same. Finally, let's look at y equals 03:06:40.369 --> 03:06:42.079 one over x squared. 03:06:42.079 --> 03:06:48.689 Again, it's not defined when x is zero, but I can plug in points like one half or X, let's 03:06:48.689 --> 03:06:55.498 see one over one half squared is one over 1/4, which is four. 03:06:55.498 --> 03:07:03.369 And one over one squared, one over two squared is a fourth, it looks pretty much like the 03:07:03.369 --> 03:07:08.399 previous function is just a little bit more extreme rises a little more steeply, falls 03:07:08.398 --> 03:07:12.959 a little more dramatically. But for negative values of x, something a little bit different 03:07:12.959 --> 03:07:19.039 goes on. For example, one over negative two squared is just one over four, which is a 03:07:19.040 --> 03:07:27.060 fourth. So I can plot that point there, and one over a negative one squared, that's just 03:07:27.059 --> 03:07:32.260 one. So my curve for negative values of x is gonna lie in the second quadrant, not the 03:07:32.260 --> 03:07:38.270 third quadrant. This is an example of an even function 03:07:38.270 --> 03:07:42.359 because it has perfect mirror symmetry across the y axis. 03:07:42.359 --> 03:07:49.809 These are the toolkit functions, and I recommend you memorize the shape of each of them. 03:07:49.809 --> 03:07:56.180 That way, you can draw at least a rough sketch very quickly without having to plug in points. 03:07:56.180 --> 03:07:58.850 That's all for the graphs of the toolkit functions. 03:07:58.850 --> 03:08:07.029 If we change the equation of a function, then the graph of the function changes or transforms 03:08:07.029 --> 03:08:09.459 in predictable ways. 03:08:09.459 --> 03:08:14.460 This video gives some rules and examples for transformations of functions. To get the most 03:08:14.459 --> 03:08:19.108 out of this video, it's helpful if you're already familiar with the graph of some typical 03:08:19.109 --> 03:08:25.439 functions, I call them toolkit functions like y equals square root of x, y equals x squared, 03:08:25.439 --> 03:08:30.350 y equals the absolute value of x and so on. If you're not familiar with those graphs, 03:08:30.350 --> 03:08:36.930 I encourage you to watch my video called toolkit functions first, before watching this one. 03:08:36.930 --> 03:08:39.520 I want to start by reviewing function notation. 03:08:39.520 --> 03:08:46.340 If g of x represents the function, the square root of x, then we can rewrite these expressions 03:08:46.340 --> 03:08:53.199 in terms of square roots. For example, g of x minus two is the same thing as the square 03:08:53.199 --> 03:08:55.569 root of x minus two. 03:08:55.568 --> 03:09:05.889 g of quantity x minus two means we plug in x minus two, everywhere we see an x, so that 03:09:05.889 --> 03:09:11.978 would be the same thing as the square root of quantity x minus two. 03:09:11.978 --> 03:09:16.840 In this second example, I say that we're subtracting two on the inside of the function, because 03:09:16.840 --> 03:09:21.930 we're subtracting two before we apply the square root function. Whereas on the first 03:09:21.930 --> 03:09:26.990 example, I say that the minus two is on the outside of the function, we're doing the square 03:09:26.990 --> 03:09:33.359 root function first and then subtracting two. In this third example, g of 3x, we're multiplying 03:09:33.359 --> 03:09:38.590 by three on the inside of the function. To evaluate this in terms of square root, we 03:09:38.590 --> 03:09:44.549 plug in the entire 3x for x and the square root function. That gives us the square root 03:09:44.549 --> 03:09:45.568 of 3x. 03:09:45.568 --> 03:09:51.609 In the next example, we're multiplying by three on the outside of the function. This 03:09:51.609 --> 03:09:58.720 is just three times the square root of x. Finally, g of minus x means the square root 03:09:58.719 --> 03:10:00.059 of minus x. 03:10:00.059 --> 03:10:03.769 Now, this might look a little odd because we're not used to taking the square root of 03:10:03.770 --> 03:10:09.760 a negative number. But remember that if x itself is negative, like negative to the negative 03:10:09.760 --> 03:10:13.850 x will be negative negative two or positive two. So we're really be taking the square 03:10:13.850 --> 03:10:18.899 root of a positive number in that case, let me record which of these are inside and which 03:10:18.898 --> 03:10:21.748 of these are outside of my function. 03:10:21.748 --> 03:10:26.408 In this next set of examples, we're using the same function g of x squared of x. But 03:10:26.408 --> 03:10:29.958 this time, we're starting with an expression with square roots in it, and trying to write 03:10:29.959 --> 03:10:37.699 it in terms of g of x. So the first example, the plus 17 is on the outside of my function, 03:10:37.699 --> 03:10:43.248 because I'm taking the square root of x first, and then adding 17. So I can write this as 03:10:43.248 --> 03:10:46.418 g of x plus 17. 03:10:46.418 --> 03:10:53.520 In the second example, I'm taking x and adding 12 first, then I take the square root of the 03:10:53.520 --> 03:10:58.550 whole thing. Since I'm adding the 12 to x first that's on the inside of my function. 03:10:58.549 --> 03:11:04.049 So I write that as g of the quantity x plus 12. 03:11:04.049 --> 03:11:09.559 Remember that this notation means I plug in the entire x plus 12, into the square root 03:11:09.559 --> 03:11:14.738 sign, which gives me exactly square root of x plus 12. And this next example, undoing 03:11:14.738 --> 03:11:20.539 the square root first and then multiplying by negative 36. So i minus 36, multiplications, 03:11:20.540 --> 03:11:27.760 outside my function, I can rewrite this as minus 36 times g of x. Finally, in this last 03:11:27.760 --> 03:11:33.130 example, I take x multiplied by a fourth and then apply the square root of x. So that's 03:11:33.129 --> 03:11:40.938 the same thing as g of 1/4 X, my 1/4 X is on the inside of my function. In other words, 03:11:40.939 --> 03:11:44.880 it's inside the parentheses when I use function notation. 03:11:44.879 --> 03:11:50.809 Let's graph the square root of x and two transformations have this function, 03:11:50.809 --> 03:11:56.019 y equals the square root of x goes to the point 0011. And for two, since the square 03:11:56.020 --> 03:12:01.270 root of four is two, it looks something like this. 03:12:01.270 --> 03:12:06.248 In order to graph y equals the square root of x minus two, notice that the minus two 03:12:06.248 --> 03:12:10.629 is on the outside of the function, that means we're going to take the square root of x first 03:12:10.629 --> 03:12:17.299 and then subtract two. So for example, if we start with the x value of zero, and compute 03:12:17.299 --> 03:12:21.938 the square root of zero, that's zero, then we subtract two to give us a y value of negative 03:12:21.939 --> 03:12:23.680 two, 03:12:23.680 --> 03:12:29.510 an x value of one, which under the square root function had a y value of one now has 03:12:29.510 --> 03:12:36.898 a y value that's decreased by two, one minus two is negative one. And finally, an x value 03:12:36.898 --> 03:12:42.469 of four, which under the square root function had a y value of two now has a y value of 03:12:42.469 --> 03:12:49.840 two minus two or zero, its y value is also decreased by two. If I plot these new points, 03:12:49.840 --> 03:12:56.469 zero goes with negative two, one goes with negative one, and four goes with zero, I have 03:12:56.469 --> 03:13:00.078 my transformed graph. 03:13:00.078 --> 03:13:04.978 Because I subtracted two on the outside of my function, my y values were decreased by 03:13:04.978 --> 03:13:12.118 two, which brought my graph down by two units. Next, let's look at y equals the square root 03:13:12.119 --> 03:13:17.359 of quantity x minus two. Now we're subtracting two on the inside of our function, which means 03:13:17.359 --> 03:13:21.899 we subtract two from x first and then take the square root. In order to get the same 03:13:21.898 --> 03:13:28.279 y value of zero as we had in our blue graph, we need our x minus two to be zero, so we 03:13:28.280 --> 03:13:30.430 need our original x to be two. 03:13:30.430 --> 03:13:36.689 In order to get the y value of one that we had in our blue graph, we need to be taking 03:13:36.689 --> 03:13:41.659 the square root of one, so we need x minus two to be one, which means that we need to 03:13:41.659 --> 03:13:43.459 start with an x value of three. 03:13:43.459 --> 03:13:50.819 And in order to reproduce our y value of two from our original graph, we need the square 03:13:50.819 --> 03:13:55.408 root of x minus two to be two, which means we need to start out by taking the square 03:13:55.408 --> 03:14:04.168 root of four, which means our x minus two is four, so our x should be up there at six. 03:14:04.168 --> 03:14:10.510 If I plot my x values, with my corresponding y values of square root of x minus two, I 03:14:10.510 --> 03:14:18.068 get the following graph. Notice that the graph has moved horizontally to the right by two 03:14:18.068 --> 03:14:19.068 units. 03:14:19.068 --> 03:14:25.629 To me, moving down by two units, makes sense because we're subtracting two, so we're decreasing 03:14:25.629 --> 03:14:30.519 y's by two units, but the minus two on the inside that kind of does the opposite of what 03:14:30.520 --> 03:14:35.439 I expect, I might expected to to move the graph left, I might expect the x values to 03:14:35.439 --> 03:14:41.389 be going down by two units, but instead, it moves the graph to the right, 03:14:41.389 --> 03:14:46.869 because the x units have to go up by two units in order to get the right square root when 03:14:46.870 --> 03:14:52.919 I then subtract two units again, the observations we made for these transformations of functions 03:14:52.918 --> 03:14:58.590 hold in general, according to the following rules. First of all, numbers on the outside 03:14:58.590 --> 03:14:59.850 of the function like 03:14:59.850 --> 03:15:06.399 In our example, y equals a squared of x minus two, those numbers affect the y values. And 03:15:06.398 --> 03:15:12.148 a result in vertical motions, like we saw, these motions are in the direction you expect. 03:15:12.148 --> 03:15:18.119 So subtracting two was just down by two. If we were adding two instead, that would move 03:15:18.120 --> 03:15:19.880 us up by two 03:15:19.879 --> 03:15:24.829 numbers on the inside of the function. That's like our example, y equals the square root 03:15:24.829 --> 03:15:31.379 of quantity x minus two, those affect the x values and results in a horizontal motion, 03:15:31.379 --> 03:15:35.858 these motions go in the opposite direction from what you expect. Remember, the minus 03:15:35.859 --> 03:15:40.658 two on the inside actually shifted our graph to the right, if it had had a plus two on 03:15:40.658 --> 03:15:45.918 the inside, that would actually shift our graph to the left. 03:15:45.918 --> 03:15:51.738 Adding results in a shift those are called translations, but multiplying like something 03:15:51.738 --> 03:15:59.568 like y equals three times the square root of x, that would result in a stretch or shrink. 03:15:59.568 --> 03:16:03.748 In other words, if I start with the square root of x, 03:16:03.748 --> 03:16:08.539 and then when I graph y equals three times the square root of x, that stretches my graph 03:16:08.539 --> 03:16:12.590 vertically by a factor of three. 03:16:12.590 --> 03:16:13.590 Like this, 03:16:13.590 --> 03:16:20.789 if I want to graph y equals 1/3, times the square root of x, that shrinks my graph vertically 03:16:20.789 --> 03:16:22.909 by a factor of 1/3. 03:16:22.908 --> 03:16:29.469 Finally, a negative sign results in reflection. For example, if I start with the graph of 03:16:29.469 --> 03:16:34.750 y equals the square root of x, and then when I graph y equals the square root of negative 03:16:34.750 --> 03:16:39.859 x, that's going to do a reflection in the horizontal direction because the negative 03:16:39.859 --> 03:16:42.130 is on the inside of the square root sign. 03:16:42.129 --> 03:16:49.318 A reflection in the horizontal direction means a reflection across the y axis. 03:16:49.318 --> 03:16:53.939 If instead, I want to graph y equals negative A squared of x, that negative sign on the 03:16:53.939 --> 03:17:00.039 outside means a vertical reflection, a reflection across the x axis. 03:17:00.039 --> 03:17:06.529 Pause the video for a moment and see if you could describe what happens in these four 03:17:06.529 --> 03:17:07.779 transformations. 03:17:07.779 --> 03:17:12.859 In the first example, we're subtracting four on the outside of the function. adding or 03:17:12.859 --> 03:17:18.380 subtracting means a translation or shift. And since we're on the outside of the function 03:17:18.379 --> 03:17:24.049 affects the y value, so that's moving us vertically. So this transformation should take the square 03:17:24.049 --> 03:17:31.358 root of graph and move it down by four units, that would look something like this. 03:17:31.359 --> 03:17:39.090 In the next example, we're adding 12 on the inside, that's still a translation. But now 03:17:39.090 --> 03:17:44.719 we're moving horizontally. And so since we go the opposite direction, we expect we are 03:17:44.719 --> 03:17:51.148 going to go to the left by 12 units, that's going to look something like this. 03:17:51.148 --> 03:17:55.278 And the next example, we're multiplying by three and introducing a negative sign both 03:17:55.279 --> 03:18:00.689 on the outside of our function outside our function means we're affecting the y values. 03:18:00.689 --> 03:18:06.120 So in multiplication means we're stretching by a factor of three, the negative sign means 03:18:06.120 --> 03:18:10.310 we reflect in the vertical direction, 03:18:10.309 --> 03:18:15.168 here's stretching by a factor of three vertically, before I apply the minus sign, and now the 03:18:15.168 --> 03:18:19.020 minus sign reflects in the vertical direction. 03:18:19.020 --> 03:18:26.029 Finally, in this last example, we're multiplying by one quarter on the inside of our function, 03:18:26.029 --> 03:18:31.040 we know that multiplication means stretch or shrink. And since we're on the inside, 03:18:31.040 --> 03:18:36.220 it's a horizontal motion, and it does the opposite of what we expect. So instead of 03:18:36.219 --> 03:18:41.648 shrinking by a factor of 1/4, horizontally, it's actually going to stretch by the reciprocal, 03:18:41.648 --> 03:18:44.139 a factor of four horizontally. 03:18:44.139 --> 03:18:50.998 that'll look something like this. Notice that stretching horizontally by a factor of four 03:18:50.998 --> 03:18:55.689 looks kind of like shrinking vertically by a factor of one half. 03:18:55.689 --> 03:19:01.000 And that's actually borne out by the algebra, because the square root of 1/4 x is the same 03:19:01.000 --> 03:19:05.510 thing as the square root of 1/4 times the square root of x, which is the same thing 03:19:05.510 --> 03:19:12.119 as one half times the square root of x. And so now we can see algebraically that vertical 03:19:12.119 --> 03:19:18.520 shrink by a factor of one half is the same as a horizontal stretch by a factor of four, 03:19:18.520 --> 03:19:22.310 at least for this function, the square root function. 03:19:22.309 --> 03:19:27.988 This video gives some rules for transformations of functions, which I'll repeat below. numbers 03:19:27.988 --> 03:19:37.039 on the outside correspond to changes in the y values, or vertical motions. 03:19:37.040 --> 03:19:44.439 numbers on the inside of the function, affect the x values, and result in horizontal motions. 03:19:44.439 --> 03:19:47.760 Adding and subtracting 03:19:47.760 --> 03:19:52.020 corresponds to translations or shifts. 03:19:52.020 --> 03:19:57.988 multiplying and dividing by numbers corresponds to stretches and shrinks 03:19:57.988 --> 03:20:01.389 and putting in a negative sign. 03:20:01.389 --> 03:20:04.269 corresponds to a reflection, 03:20:04.270 --> 03:20:10.100 horizontal reflection, if the negative sign is on the inside, and a vertical reflection, 03:20:10.100 --> 03:20:13.010 if the negative sign is on the outside, 03:20:13.010 --> 03:20:17.939 knowing these basic rules about transformations empowers you to be able to sketch graphs of 03:20:17.939 --> 03:20:24.040 much more complicated functions, like y equals three times the square root of x plus two, 03:20:24.040 --> 03:20:29.521 by simply considering the transformations, one at a time. 03:20:29.521 --> 03:20:35.770 a quadratic function is a function that can be written in the form f of x equals A x squared 03:20:35.770 --> 03:20:46.918 plus bx plus c, where a, b and c are real numbers. And a is not equal to zero. The reason 03:20:46.918 --> 03:20:53.449 we require that A is not equal to zero is because if a were equal to zero, we'd have 03:20:53.449 --> 03:21:01.449 f of x is equal to b x plus c, which is called a linear function. So by making sure that 03:21:01.449 --> 03:21:06.459 a is not zero, we make sure there's really an x squared term which is the hallmark of 03:21:06.459 --> 03:21:08.959 a quadratic function. 03:21:08.959 --> 03:21:13.939 Please pause the video for a moment and decide which of these equations represent quadratic 03:21:13.939 --> 03:21:15.869 functions. 03:21:15.869 --> 03:21:23.110 The first function can definitely be written in the form g of x equals A x squared plus 03:21:23.110 --> 03:21:30.840 bx plus c. fact it's already written in that form, where a is negative five, B is 10, and 03:21:30.840 --> 03:21:32.199 C is three. 03:21:32.199 --> 03:21:39.859 The second equation is also a quadratic function, because we can think of it as f of x equals 03:21:39.859 --> 03:21:46.559 one times x squared plus zero times x plus zero. 03:21:46.559 --> 03:21:53.379 So it is in the right form, where A is one, B is zero, and C is zero. 03:21:53.379 --> 03:21:59.108 It's perfectly fine for the coefficient of x and the constant term to be zero for a quadratic 03:21:59.109 --> 03:22:03.590 function, we just need the coefficient of x squared to be nonzero, so the x squared 03:22:03.590 --> 03:22:06.380 term is preserved. 03:22:06.379 --> 03:22:13.269 The third equation is not a quadratic function. It's a linear function, because there's no 03:22:13.270 --> 03:22:17.069 x squared term. 03:22:17.068 --> 03:22:21.998 The fourth function might not look like a quadratic function. But if we rewrite it by 03:22:21.998 --> 03:22:28.708 expanding out the X minus three squared, let's see what happens. We get y equals two times 03:22:28.709 --> 03:22:36.760 x minus three times x minus three plus four. So that's two times x squared minus 3x minus 03:22:36.760 --> 03:22:45.228 3x, plus nine plus four, continuing, I get 2x squared, 03:22:45.228 --> 03:22:53.260 minus 12 access, plus 18 plus four, in other words, y equals 2x squared minus 12x plus 03:22:53.260 --> 03:23:02.488 22. So in fact, our function can be written in the right form, and it is a quadratic function. 03:23:02.488 --> 03:23:08.488 A function that is already written in the form y equals A x squared plus bx plus c is 03:23:08.488 --> 03:23:12.520 said to be in standard form. 03:23:12.520 --> 03:23:17.289 So our first example g of x is in standard form 03:23:17.289 --> 03:23:23.270 a function that's written in the format of the last function that is in the form of y 03:23:23.270 --> 03:23:33.529 equals a times x minus h squared plus k for some numbers, a, h, and K. That said to be 03:23:33.529 --> 03:23:35.760 in vertex form, 03:23:35.760 --> 03:23:40.498 I'll talk more about standard form and vertex form in another video. 03:23:40.498 --> 03:23:45.869 In this video, we identified some quadratic functions, and talked about the difference 03:23:45.869 --> 03:23:49.529 between standard form 03:23:49.529 --> 03:23:54.180 and vertex form. 03:23:54.180 --> 03:23:59.529 This video is about graphing quadratic functions. a quadratic function, which is typically written 03:23:59.529 --> 03:24:08.069 in standard form, like this, or sometimes in vertex form, like this, always has the 03:24:08.068 --> 03:24:11.350 graph that looks like a parabola. 03:24:11.350 --> 03:24:17.229 This video will show how to tell whether the problem is pointing up or down. How to find 03:24:17.228 --> 03:24:21.788 its x intercepts, and how to find its vertex. 03:24:21.789 --> 03:24:30.890 The bare bones basic quadratic function is f of x equals x squared, it goes to the origin, 03:24:30.889 --> 03:24:38.568 since f of zero is zero squared, which is zero, and it is a parabola pointing upwards 03:24:38.568 --> 03:24:40.738 like this. 03:24:40.738 --> 03:24:46.270 The vertex of a parabola is its lowest point if it's pointing upwards, and its highest 03:24:46.270 --> 03:24:53.908 point if it's pointing downwards. So in this case, the vertex is at 00. 03:24:53.908 --> 03:24:58.949 The x intercepts are where the graph crosses the x axis. In other words, where y is zero 03:24:58.949 --> 03:25:05.529 In this function, y equals zero means that x squared is zero, which happens only when 03:25:05.529 --> 03:25:11.350 x is zero. So the x intercept, there's only one of them is also zero. 03:25:11.350 --> 03:25:17.869 The second function, y equals negative 3x squared also goes to the origin. Since the 03:25:17.869 --> 03:25:25.140 functions value when x is zero, is y equals zero. But in this case, the parabola is pointing 03:25:25.139 --> 03:25:27.818 downwards. 03:25:27.818 --> 03:25:32.100 That's because thinking about transformations of functions, a negative sign on the outside 03:25:32.100 --> 03:25:38.600 reflects the function vertically over the x axis, making the problem instead of pointing 03:25:38.600 --> 03:25:43.819 upwards, reflecting the point downwards. 03:25:43.819 --> 03:25:49.010 The number three on the outside stretches the graph vertically by a factor of three. 03:25:49.010 --> 03:25:55.309 So it makes it kind of long and skinny like this. 03:25:55.309 --> 03:26:00.299 In general, a negative coefficient to the x squared term means the problem will be pointing 03:26:00.299 --> 03:26:05.248 down. Whereas a positive coefficient, like here, the coefficients, one means the parabola 03:26:05.248 --> 03:26:06.738 is pointing up. 03:26:06.738 --> 03:26:09.648 Alright, that roll over here. 03:26:09.648 --> 03:26:16.129 So if a is bigger than zero, the parabola opens up. 03:26:16.129 --> 03:26:22.159 And if the value of the coefficient a is less than zero, then the parabola opens down. 03:26:22.159 --> 03:26:27.728 In this second example, we can see again that the vertex is at 00. And the x intercept is 03:26:27.728 --> 03:26:30.408 x equals zero. 03:26:30.408 --> 03:26:33.538 Let's look at this third example. 03:26:33.539 --> 03:26:38.168 If we multiplied our expression out, we'd see that the coefficient of x squared would 03:26:38.168 --> 03:26:43.760 be to a positive number. So that means our parabola is going to be opening up. 03:26:43.760 --> 03:26:50.020 But the vertex of this parabola will no longer be at the origin. In fact, we can find the 03:26:50.020 --> 03:26:55.319 parabolas vertex by thinking about transformations of functions. 03:26:55.318 --> 03:27:02.318 Our function is related to the function y equals to x squared, by moving it to the right 03:27:02.318 --> 03:27:13.889 by three and up by four. Since y equals 2x squared has a vertex at 00. If we move that 03:27:13.889 --> 03:27:21.549 whole parabola including the vertex, right by three, and up by four, the vertex will 03:27:21.549 --> 03:27:26.099 end up at the point three, four. 03:27:26.100 --> 03:27:30.170 So a parabola will look something like this. 03:27:30.170 --> 03:27:36.680 Notice how easy it was to just read off the vertex when our quadratic function is written 03:27:36.680 --> 03:27:44.050 in this form. In fact, any parabola any quadratic function written in the form a times x minus 03:27:44.049 --> 03:27:53.269 h squared plus k has a vertex at h k. By the same reasoning, we're moving the parabola 03:27:53.270 --> 03:27:59.779 with a vertex at the origin to the right by H, and by K. 03:27:59.779 --> 03:28:06.300 That's why this form of a quadratic function is called the vertex form. 03:28:06.299 --> 03:28:11.879 Notice that this parabola has no x intercepts because it does not cross the x axis. 03:28:11.879 --> 03:28:19.118 For our final function, we have g of x equals 5x squared plus 10x plus three, 03:28:19.119 --> 03:28:23.590 we know the graph of this function will be a parabola pointing upwards, because the coefficient 03:28:23.590 --> 03:28:26.139 of x squared is positive. 03:28:26.139 --> 03:28:32.519 To find the x intercepts, we can set y equals zero, since the x intercepts is where our 03:28:32.520 --> 03:28:37.229 graph crosses the x axis, and that's where the y value is zero. 03:28:37.228 --> 03:28:43.858 So zero equals 5x squared plus 10x plus three, I'm going to use the quadratic formula to 03:28:43.859 --> 03:28:52.029 solve that. So x is negative 10 plus or minus the square root of 10 squared minus four times 03:28:52.029 --> 03:28:59.748 five times three, all over two times five. That simplifies to x equals negative 10 plus 03:28:59.748 --> 03:29:03.590 or minus the square root of 40 over 10, 03:29:03.590 --> 03:29:08.139 which simplifies further to x equals negative 10 over 10, plus or minus the square root 03:29:08.139 --> 03:29:16.478 of 40 over 10, which is negative one plus or minus two square root of 10 over 10, or 03:29:16.478 --> 03:29:21.288 negative one plus or minus square root of 10 over five. 03:29:21.289 --> 03:29:27.409 Since the square root of 10 is just a little bit bigger than three, this works out to approximately 03:29:27.408 --> 03:29:34.359 about negative two fifths and negative eight foot or so somewhere around right here. So 03:29:34.359 --> 03:29:40.600 our parabola is going to look something like this. Notice that it goes through crosses 03:29:40.600 --> 03:29:48.168 the x axis at y equals three, that's because when we plug in x equals zero, and two here 03:29:48.168 --> 03:29:56.969 we get y equal three, so the y intercept is at three. Finally, we can find the vertex. 03:29:56.969 --> 03:30:00.299 Since this function is written in standard form, 03:30:00.299 --> 03:30:10.208 On the Y equals a x plus a squared plus bx plus c form, and not in vertex form, we can't 03:30:10.209 --> 03:30:15.120 just read off the vertex like we couldn't before. But there's a trick called the vertex 03:30:15.120 --> 03:30:20.279 formula, which says that whenever you have a function in a quadratic function in standard 03:30:20.279 --> 03:30:24.040 form, the vertex has an x coordinate 03:30:24.040 --> 03:30:34.090 of negative B over two A. So in this case, that's an x coordinate of negative 10 over 03:30:34.090 --> 03:30:39.988 two times five or negative one, which is kind of like where I put it on the graph. To find 03:30:39.988 --> 03:30:47.578 the y coordinate of that vertex, I can just plug in negative one into my equation for 03:30:47.578 --> 03:30:54.799 x, which gives me at y equals five times negative one squared, plus 10, times negative one plus 03:30:54.799 --> 03:30:59.420 three, which is negative two. 03:30:59.420 --> 03:31:04.398 So I think I better redraw my graph a little bit to put that vertex down at a y coordinate 03:31:04.398 --> 03:31:10.939 of negative two where it's supposed to be. 03:31:10.939 --> 03:31:15.828 Let's summarize the steps we use to graph these quadratic functions. First of all, the 03:31:15.828 --> 03:31:19.680 graph of a quadratic function has the shape of a parabola. 03:31:19.680 --> 03:31:27.318 The parabola opens up, if the coefficient of x squared, which we'll call a is greater 03:31:27.318 --> 03:31:34.658 than zero and down if a is less than zero. To find the x intercepts, we set y equal to 03:31:34.658 --> 03:31:44.199 zero, or in other words, f of x equal to zero and solve for x, the find the vertex, we can 03:31:44.199 --> 03:31:53.310 either read it off as h k, if our function is in vertex form, 03:31:53.309 --> 03:31:57.189 or we can use the vertex formula 03:31:57.189 --> 03:32:01.260 and get the x coordinate 03:32:01.260 --> 03:32:10.930 of the vertex to be negative B over to a if our function is in standard form. 03:32:10.930 --> 03:32:19.520 To find the y coordinate of the vertex in this case, we just plug in the x coordinate 03:32:19.520 --> 03:32:21.789 and figure out what y is. 03:32:21.789 --> 03:32:29.590 Finally, we can always find additional points on the graph by plugging in values of x. 03:32:29.590 --> 03:32:34.870 In this video, we learned some tricks for graphing quadratic functions. In particular, 03:32:34.870 --> 03:32:43.220 we saw that the vertex can be read off as h k, if our function is written in vertex 03:32:43.219 --> 03:32:50.478 form, and the x coordinate of the vertex can be calculated 03:32:50.478 --> 03:32:57.198 as negative B over two A, if our function is in standard form. For an explanation of 03:32:57.199 --> 03:33:02.959 why this vertex formula works, please see the my other video. 03:33:02.959 --> 03:33:09.920 a quadratic and standard form looks like y equals A x squared plus bx plus c, where a, 03:33:09.920 --> 03:33:16.568 b and c are real numbers, and a is not zero. a quadratic function in vertex form looks 03:33:16.568 --> 03:33:24.028 like y equals a times x minus h squared plus k, where a h and k are real numbers, and a 03:33:24.029 --> 03:33:30.328 is not zero. When a functions in vertex form, it's easy to read off the vertex is just the 03:33:30.328 --> 03:33:34.488 ordered pair, H, okay. 03:33:34.488 --> 03:33:40.568 This video explains how to get from vertex form two standard form and vice versa. 03:33:40.568 --> 03:33:46.549 Let's start by converting this quadratic function from vertex form to standard form. That's 03:33:46.549 --> 03:33:50.568 pretty straightforward, we just have to distribute out. 03:33:50.568 --> 03:33:58.668 So if I multiply out the X minus three squared, 03:33:58.668 --> 03:34:07.289 I get minus four times x squared minus 6x plus nine plus one, distributing the negative 03:34:07.289 --> 03:34:16.918 four, I get negative 4x squared plus 24x minus 36 plus one. So that works out to minus 4x 03:34:16.918 --> 03:34:26.709 squared plus 24x minus 35. And I have my quadratic function now in standard form. 03:34:26.709 --> 03:34:31.618 Now let's go the other direction and convert a quadratic function that's already in standard 03:34:31.619 --> 03:34:40.510 form into vertex form. That is, we want to put it in the form of g of x equals a times 03:34:40.510 --> 03:34:51.090 x minus h squared plus k, where the vertex is at h k. To do this, it's handy to use the 03:34:51.090 --> 03:34:53.870 vertex formula. 03:34:53.870 --> 03:35:00.380 The vertex formula says that the x coordinate of the vertex is given by negative 03:35:00.379 --> 03:35:08.368 over to a, where A is the coefficient of x squared, and B is the coefficient of x. So 03:35:08.369 --> 03:35:16.649 in this case, we get an x coordinate of negative eight over two times two or negative two. 03:35:16.648 --> 03:35:24.118 To find the y coordinate of the vertex, we just plug in the x coordinate into our formula 03:35:24.119 --> 03:35:30.918 for a g of x. So that's g of negative two, which is two times negative two squared plus 03:35:30.918 --> 03:35:37.360 eight times negative two plus six. And that works out to be negative two by coincidence. 03:35:37.360 --> 03:35:44.461 So the vertex for our quadratic function has coordinates negative two, negative two. And 03:35:44.460 --> 03:35:50.429 if I want to write g of x in vertex form, it's going to be a times x minus negative 03:35:50.430 --> 03:36:00.789 two squared plus minus two. That's because remember, we subtract H and we add K. So that 03:36:00.789 --> 03:36:08.699 simplifies to g of x equals a times x plus two squared minus two. And finally, we just 03:36:08.699 --> 03:36:14.359 need to figure out what this leading coefficient as. But notice, if we were to multiply, distribute 03:36:14.359 --> 03:36:23.029 this out, then the coefficient of x squared would end up being a. So therefore, the coefficient 03:36:23.029 --> 03:36:27.520 of x squared here, which is a has to be the same as the coefficient of x squared here, 03:36:27.520 --> 03:36:33.149 which we conveniently also called a, in other words, are a down here needs to be two. So 03:36:33.148 --> 03:36:39.778 I'm going to write that as g of x equals two times x plus two squared minus two, lots of 03:36:39.779 --> 03:36:46.289 twos in this problem. And that's our quadratic function in vertex form. If I want to check 03:36:46.289 --> 03:36:53.569 my answer, of course, I could just distribute out again, I'd get two times x squared plus 03:36:53.568 --> 03:37:03.000 4x, plus two, minus two, in other words, 2x squared plus 8x plus six, which checks out 03:37:03.000 --> 03:37:09.010 to exactly what I started with. This video showed how to get from vertex form to standard 03:37:09.010 --> 03:37:14.408 form by distributing out and how to get from standard form to vertex form by finding the 03:37:14.408 --> 03:37:17.639 vertex using the vertex formula. 03:37:17.639 --> 03:37:26.269 Suppose you have a quadratic function in the form y equals x squared plus bx plus c, and 03:37:26.270 --> 03:37:31.760 you want to find where the vertex is, when you graph it. 03:37:31.760 --> 03:37:40.969 The vertex formula says that the x coordinate of this vertex is that negative B over two 03:37:40.969 --> 03:37:42.879 A, 03:37:42.879 --> 03:37:47.608 this video gives a justification for where that formula comes from. 03:37:47.609 --> 03:37:53.390 Let's start with a specific example. Suppose I wanted to find the x intercepts and the 03:37:53.389 --> 03:38:00.840 vertex for this quadratic function. To find the x intercepts, I would set y equal to zero 03:38:00.840 --> 03:38:07.510 and solve for x. So that's zero equals 3x squared plus 7x minus five. And to solve for 03:38:07.510 --> 03:38:13.418 x, I use the quadratic formula. So x is going to be negative seven plus or minus the square 03:38:13.418 --> 03:38:22.328 root of seven squared minus four times three times negative five, all over two times three. 03:38:22.328 --> 03:38:29.228 That simplifies to negative seven plus or minus a squared of 109 over six, we could, 03:38:29.228 --> 03:38:35.039 I could also write this as negative seven, six, plus the square root of 109 over six, 03:38:35.040 --> 03:38:41.029 or x equals negative seven, six minus the square root of 109 over six, since the square 03:38:41.029 --> 03:38:47.488 root of 109 is just a little bit bigger than 10, this can be approximated by negative seven 03:38:47.488 --> 03:38:54.340 six plus 10, six, and negative seven, six minus 10, six. 03:38:54.340 --> 03:39:00.709 So pretty close to, I guess about what half over here and pretty close to about negative 03:39:00.709 --> 03:39:06.609 three over here, I'm just going to estimate so that I can draw a picture of the function. 03:39:06.609 --> 03:39:11.629 Since the leading coefficient three is positive, I know my parabola is going to be opening 03:39:11.629 --> 03:39:20.270 up and the intercepts are somewhere around here and here. So roughly speaking, it's gonna 03:39:20.270 --> 03:39:25.488 look something like this. 03:39:25.488 --> 03:39:30.439 Now the vertex is going to be somewhere in between the 2x intercepts, in fact, it's going 03:39:30.439 --> 03:39:37.010 to be by symmetry, it'll be exactly halfway in between the 2x intercepts. Since the x 03:39:37.010 --> 03:39:43.260 intercepts are negative seven, six plus and minus 100 squared of 109 over six, the number 03:39:43.260 --> 03:39:47.978 halfway in between those is going to be exactly negative seven, six, 03:39:47.978 --> 03:39:56.049 right, because on the one hand, I have negative seven six plus something. And on the other 03:39:56.049 --> 03:39:59.858 hand, I have negative seven six minus that same thing. 03:39:59.859 --> 03:40:08.998 So negative seven, six will be exactly in the middle. So my x coordinate of my vertex 03:40:08.998 --> 03:40:16.670 will be at negative seven sex. Notice that I got that number from the quadratic formula. 03:40:16.670 --> 03:40:24.818 More generally, if I want to find the x intercepts for any quadratic function, I set y equal 03:40:24.818 --> 03:40:27.469 to zero 03:40:27.469 --> 03:40:33.198 and solve for x using the quadratic formula, negative b plus or minus the square root of 03:40:33.199 --> 03:40:40.890 b squared minus four AC Oliver to a, b, x intercepts will be at these two values, but 03:40:40.889 --> 03:40:47.189 the x coordinate of the vertex, which is exactly halfway in between the 2x intercepts will 03:40:47.189 --> 03:40:53.710 be at negative B over two A. That's where the vertex formula comes from. 03:40:53.709 --> 03:40:58.628 And it turns out that this formula works even when there are no x intercepts, even when 03:40:58.629 --> 03:41:04.010 the quadratic formula gives us no solutions. That vertex still has the x coordinate, negative 03:41:04.010 --> 03:41:06.658 B over two A. 03:41:06.658 --> 03:41:11.498 And that's the justification of the vertex formula. 03:41:11.498 --> 03:41:14.898 This video is about polynomials and their graphs. 03:41:14.898 --> 03:41:20.379 We call that a polynomial is a function like this one, for example. 03:41:20.379 --> 03:41:28.708 His terms are numbers times powers of x. I'll start with some definitions. The degree of 03:41:28.709 --> 03:41:35.899 a polynomial is the largest exponent. For example, for this polynomial, the degree is 03:41:35.898 --> 03:41:37.849 for 03:41:37.850 --> 03:41:44.129 the leading term is the term with the largest exponent. In the same example, the leading 03:41:44.129 --> 03:41:51.350 term is 5x to the fourth, it's conventional to write the polynomial in descending order 03:41:51.350 --> 03:41:56.418 of powers of x. So the leading term is first. But the leading term doesn't have to be the 03:41:56.418 --> 03:42:04.498 first term. If I wrote the same polynomial as y equals, say, two minus 17x minus 21x 03:42:04.498 --> 03:42:11.658 cubed plus 5x. Fourth, the leading term would still be the 5x to the fourth, even though 03:42:11.658 --> 03:42:13.260 it was last. 03:42:13.260 --> 03:42:20.029 The leading coefficient is the number in the leading term. In this example, the leading 03:42:20.029 --> 03:42:21.459 coefficient is five. 03:42:21.459 --> 03:42:28.140 Finally, the constant term is the term with no x's in it. In our same example, the constant 03:42:28.139 --> 03:42:29.139 term is to 03:42:29.139 --> 03:42:34.760 please pause the video for a moment and take a look at this next example of polynomial 03:42:34.760 --> 03:42:40.639 figure out what's the degree the leading term, the leading coefficient and the constant term. 03:42:40.639 --> 03:42:48.760 The degree is again, four, since that's the highest power we have, and the leading term 03:42:48.760 --> 03:42:56.340 is negative 7x. to the fourth, the leading coefficient is negative seven, and the constant 03:42:56.340 --> 03:42:58.148 term is 18. 03:42:58.148 --> 03:43:04.510 In the graph of the polynomial Shown here are the three marks points are called turning 03:43:04.510 --> 03:43:12.469 points, because the polynomial turns around and changes direction at those three points, 03:43:12.469 --> 03:43:18.868 those same points can also be called local extreme points, or they can be called local 03:43:18.869 --> 03:43:25.710 maximum and minimum points. For this polynomial, the degree is four, and the number of turning 03:43:25.709 --> 03:43:28.068 points is three. 03:43:28.068 --> 03:43:33.129 Let's compare the degree and the number of turning points for these next three examples. 03:43:33.129 --> 03:43:39.698 For the first one, the degree is to and there's one turning point. 03:43:39.699 --> 03:43:50.090 For the second example, that agree, is three, and there's two turning points. 03:43:50.090 --> 03:43:56.920 And for this last example, the degree is four, and there's one turning point. 03:43:56.920 --> 03:44:01.699 For this first example, and the next two, the number of turning points is exactly one 03:44:01.699 --> 03:44:07.869 less than the degree. So you might conjecture that this is always true. But in fact, this 03:44:07.869 --> 03:44:13.689 is not always true. In this last example, the degree is four, but the number of turning 03:44:13.689 --> 03:44:16.909 points is one, not three. 03:44:16.909 --> 03:44:21.510 In fact, it turns out that while the number of turning points is not always equal to a 03:44:21.510 --> 03:44:28.629 minus one, it is always less than or equal to the degree minus one. That's a useful factor. 03:44:28.629 --> 03:44:33.680 Remember, when you're sketching graphs are recognizing graphs of polynomials. 03:44:33.680 --> 03:44:39.659 The end behavior of a function is how the ends of the function look, as x gets bigger 03:44:39.659 --> 03:44:45.248 and bigger heads towards infinity, or x gets goes through larger and larger negative numbers 03:44:45.248 --> 03:44:47.859 towards negative infinity. 03:44:47.859 --> 03:44:54.818 In this first example, the graph of the function goes down as x gets towards infinity as x 03:44:54.818 --> 03:44:59.959 goes towards negative infinity. I can draw this with two little arrows pointing 03:44:59.959 --> 03:45:05.958 Down on either side, or I can say in words, that the function is falling as we had left 03:45:05.959 --> 03:45:09.459 and falling also as we had right. 03:45:09.459 --> 03:45:16.600 In the second example, the graph rises to the left and rises to the right. In the third 03:45:16.600 --> 03:45:23.379 example, the graph falls to the left, but rises to the right. And in the fourth example, 03:45:23.379 --> 03:45:29.489 it rises to the left and falls to the right. If you study these examples, and others, you 03:45:29.488 --> 03:45:36.189 might notice there's a relationship between the degree of the polynomial the leading coefficients 03:45:36.189 --> 03:45:44.090 of the polynomials and the end behavior. Specifically, these four types of end behavior are determined 03:45:44.090 --> 03:45:51.728 by whether the degree is even or odd. And by whether the leading coefficient is positive 03:45:51.728 --> 03:45:54.878 or negative. 03:45:54.879 --> 03:46:00.729 When the degree is even, and the leading coefficient is positive, like in this example, where the 03:46:00.728 --> 03:46:08.458 leading coefficient is one, we have this sorts of n behavior rising on both sides. 03:46:08.459 --> 03:46:14.069 When the degree is even at the leading coefficient is negative, like in this example, we have 03:46:14.068 --> 03:46:18.378 the end behavior that's falling on both sides 03:46:18.379 --> 03:46:23.810 when the degree is odd, and the leading coefficient is positive, that's like this example, with 03:46:23.809 --> 03:46:30.699 the degree three and the leading coefficient three, then we have this sort of end behavior. 03:46:30.700 --> 03:46:35.979 And finally, when the degree is odd, and the leading coefficient is negative, like in this 03:46:35.978 --> 03:46:40.628 example, we have this sort of NBA havior. 03:46:40.629 --> 03:46:48.409 I like to remember this chart just by thinking of the most simple examples, y equals x squared, 03:46:48.408 --> 03:46:55.079 y equals negative x squared, y equals x cubed, and y equals negative x cubed. Because I know 03:46:55.079 --> 03:47:00.209 by heart what those four examples look like, 03:47:00.209 --> 03:47:08.009 then I just have to remember that any polynomial with a even degree and positive leading coefficient 03:47:08.010 --> 03:47:11.950 has the same end behavior as x squared. 03:47:11.950 --> 03:47:18.020 And similarly, any polynomial with even degree and negative leading coefficient has the same 03:47:18.020 --> 03:47:25.729 end behavior as negative x squared. And similar statements for x cubed and negative x cubed. 03:47:25.728 --> 03:47:30.738 We can use facts about turning points and behavior, to say something about the equation 03:47:30.738 --> 03:47:34.270 of a polynomial just by looking at this graph. 03:47:34.270 --> 03:47:42.699 In this example, because of the end behavior, we know that the degree is odd, 03:47:42.699 --> 03:47:46.569 we know that the leading coefficient 03:47:46.568 --> 03:47:49.389 must be positive. 03:47:49.389 --> 03:47:58.099 And finally, since there are 1234, turning points, we know that the degree is greater 03:47:58.100 --> 03:48:01.300 than or equal to five. 03:48:01.299 --> 03:48:06.288 That's because the number of turning points is less than or equal to the degree minus 03:48:06.289 --> 03:48:10.529 one. And in this case, the number of turning points we said was four. 03:48:10.529 --> 03:48:18.238 And so solving that inequality, we get the degree is bigger than equal to five. 03:48:18.238 --> 03:48:23.129 Put in some of that information together, we see the degree could be 03:48:23.129 --> 03:48:31.059 five, or seven, or nine, or any odd number greater than or equal to five. But it couldn't 03:48:31.059 --> 03:48:38.939 be for example, three or six. Because even numbers and and also numbers less than five 03:48:38.939 --> 03:48:41.129 are right out. 03:48:41.129 --> 03:48:46.500 This video gave a lot of definitions, including the definition of degree 03:48:46.500 --> 03:48:49.168 leading coefficients, 03:48:49.168 --> 03:48:51.840 turning points, 03:48:51.840 --> 03:48:54.930 and and behavior. 03:48:54.930 --> 03:48:59.908 We saw that knowing the degree and the leading coefficient can help you make predictions 03:48:59.908 --> 03:49:06.189 about the number of turning points and the end behavior, as well as vice versa. 03:49:06.189 --> 03:49:11.068 This video is about exponential functions and their graphs. 03:49:11.068 --> 03:49:16.908 an exponential function is a function that can be written in the form f of x equals a 03:49:16.908 --> 03:49:24.420 times b to the x, where a and b are any real numbers. As long as a is not zero, and B is 03:49:24.420 --> 03:49:26.549 positive. 03:49:26.549 --> 03:49:33.170 It's important to notice that for an exponential function, the variable x is in the exponent. 03:49:33.170 --> 03:49:37.609 This is different from many other functions we've seen, for example, quite a quadratic 03:49:37.609 --> 03:49:44.309 function like f of x equals 3x squared has the variable in the base, so it's not an exponential 03:49:44.309 --> 03:49:45.500 function. 03:49:45.500 --> 03:49:53.010 For exponential functions, f of x equals a times b dx. We require that A is not equal 03:49:53.010 --> 03:50:00.100 to zero, because otherwise, we would have f of x equals zero times b dx. 03:50:00.100 --> 03:50:04.850 Which just means that f of x equals zero. And this is called a constant function, not 03:50:04.850 --> 03:50:11.149 an exponential function. Because f of x is always equal to zero 03:50:11.148 --> 03:50:17.389 in an exponential function, but we require that B is bigger than zero, because otherwise, 03:50:17.389 --> 03:50:27.578 for example, if b is equal to negative one, say, we'd have f of x equals a times negative 03:50:27.578 --> 03:50:34.038 one to the x. Now, this would make sense for a lot of values of x. But if we try to do 03:50:34.039 --> 03:50:40.600 something like f of one half, with our Bs, negative one, that would be the same thing 03:50:40.600 --> 03:50:45.340 as a times the square root of negative one, which is not a real number, 03:50:45.340 --> 03:50:52.158 we'd get the same problem for other values of b, if the values of b were negative. And 03:50:52.158 --> 03:50:59.299 if we tried b equals zero, we'd get a kind of ridiculous thing a times zero to the x, 03:50:59.299 --> 03:51:03.828 which again is always zero. So that wouldn't count as an exponential either. So we can't 03:51:03.828 --> 03:51:09.728 use any negative basis, and we can't use zero basis, if we want an exponential function. 03:51:09.728 --> 03:51:18.448 The number A in the expression f of x equals a times b to the x is called the initial value. 03:51:18.449 --> 03:51:21.029 And the number B is called the base. 03:51:21.029 --> 03:51:28.998 The phrase initial value comes from the fact that if we plug in x equals zero, we get a 03:51:28.998 --> 03:51:34.529 times b to the zero, well, anything to the zero is just one. So this is equal to A. In 03:51:34.529 --> 03:51:40.920 other words, f of zero equals a. So if we think of starting out, 03:51:40.920 --> 03:51:44.700 when x equals zero, we get the y value 03:51:44.700 --> 03:51:51.629 of a, that's why it's called the initial value. 03:51:51.629 --> 03:51:59.890 Let's start out with this example, where y equals a times b to the x next special function, 03:51:59.889 --> 03:52:06.049 and we've set a equal to one and B equals a two. Notice that the y intercept, the value 03:52:06.049 --> 03:52:09.340 when x is zero, is going to be one. 03:52:09.340 --> 03:52:20.318 If I change my a value, my initial value, the y intercept changes, the function is stretched 03:52:20.318 --> 03:52:21.318 out. 03:52:21.318 --> 03:52:27.379 If I make the value of a go to zero, and then negative, 03:52:27.379 --> 03:52:35.219 then my initial value becomes negative, and my graph flips across the x axis. Let's go 03:52:35.219 --> 03:52:41.778 back to an a value of say one, and see what happens when we change B. Right now, the B 03:52:41.779 --> 03:52:52.869 value the basis two, if I increase B, my y intercept sticks at one, but my graph becomes 03:52:52.869 --> 03:53:03.020 steeper and steeper. If I put B back down close to one, my graph becomes more flat at 03:53:03.020 --> 03:53:05.959 exactly one, my graph is just a constant. 03:53:05.959 --> 03:53:14.779 As B gets into fractional territory, point 8.7 point six, my graph starts to slope the 03:53:14.779 --> 03:53:20.119 other way, it's decreasing now instead of increasing, but notice that the y intercept 03:53:20.119 --> 03:53:26.199 still hasn't changed, I can get it more and more steep as my B gets farther and farther 03:53:26.199 --> 03:53:33.550 away from one of course, when B goes to negative territory, my graph doesn't make any sense. 03:53:33.549 --> 03:53:44.418 So a changes the y intercept, and B changes the steepness of the graph. So that it's added 03:53:44.418 --> 03:53:50.949 whether it's increasing for B values bigger than one, and decreasing for values of the 03:53:50.949 --> 03:53:55.078 less than one. 03:53:55.078 --> 03:53:58.978 we'll summarize all these observations on the next slide. 03:53:58.978 --> 03:54:04.568 We've seen that for an exponential function, y equals a times b to the x, the parameter 03:54:04.568 --> 03:54:13.159 or number a gives the y intercept, the parameter B tells us how the graph is increasing or 03:54:13.159 --> 03:54:19.889 decreasing. Specifically, if b is greater than one, the graph is increasing. 03:54:19.889 --> 03:54:23.868 And if b is less than one, the graph is decreasing. 03:54:23.869 --> 03:54:29.459 The closer B is to the number one, the flatter the graph. 03:54:29.459 --> 03:54:37.600 So for example, if I were to graph y equals point two five to the x and y equals point 03:54:37.600 --> 03:54:44.379 four to the x, they would both be decreasing graphs, since the base for both of them is 03:54:44.379 --> 03:54:52.209 less than one. But point two five is farther away from one and point four is closer to 03:54:52.209 --> 03:54:58.560 one. So point four is going to be flatter. And point two five is going to be more steep. 03:54:58.559 --> 03:55:03.418 So in this picture, This red graph would correspond to point two 03:55:03.418 --> 03:55:11.728 five to the x, and the blue graph would correspond to point four to the x. For all these exponential 03:55:11.728 --> 03:55:17.458 functions, whether the graphs are decreasing or increasing, they all have a horizontal 03:55:17.459 --> 03:55:25.460 asymptote along the x axis. In other words, at the line at y equals zero, the domain is 03:55:25.459 --> 03:55:34.799 always from negative infinity to infinity, and the range is always from zero to infinity, 03:55:34.799 --> 03:55:43.309 because the range is always positive y values. Actually, that's true. If a is greater than 03:55:43.309 --> 03:55:52.109 zero, if a is less than zero, then our graph flips over the x axis, our domain stays the 03:55:52.109 --> 03:55:59.890 same, but our range becomes negative infinity to zero. The most famous exponential function 03:55:59.889 --> 03:56:04.868 is f of x equals e to the x. This function is also sometimes written as f of x equals 03:56:04.869 --> 03:56:13.420 x of x. The number E is Oilers number as approximately 2.7. This function has important applications 03:56:13.420 --> 03:56:21.100 to calculus and to some compound interest problems. In this video, we looked at exponential 03:56:21.100 --> 03:56:28.879 functions, functions of the form a times b to the x, where the variable is in the exponent, 03:56:28.879 --> 03:56:37.720 we saw that they all have the same general shape, either increasing like this, or decreasing 03:56:37.719 --> 03:56:45.929 like this, unless a is negative, in which case they flip over the x axis. They all have 03:56:45.930 --> 03:56:55.859 a horizontal asymptote at y equals zero, the x axis. In this video, we'll use exponential 03:56:55.859 --> 03:57:02.748 functions to model real world examples. Let's suppose you're hired for a job. The starting 03:57:02.748 --> 03:57:10.039 salary is $40,000. With a guaranteed annual raise of 3% each year, how much will your 03:57:10.039 --> 03:57:16.899 salary be after one year, two years, five years. And in general after two years, let 03:57:16.898 --> 03:57:23.349 me chart out the information. The left column will be the number of years since you're hired. 03:57:23.350 --> 03:57:30.930 And the right column will be your salary. When you start work, at zero years after you're 03:57:30.930 --> 03:57:40.970 hired, your salary will be $40,000. After one year, you'll have gotten a 3% raise. So 03:57:40.969 --> 03:57:52.459 your salary will be the original 40,000 plus 3% of 40,000.03 times 40,000. I can think 03:57:52.459 --> 03:58:00.528 of this first number as one at times 40,000. And I can factor out the 40,000 from both 03:58:00.529 --> 03:58:15.659 terms, to get 40,000 times one plus 0.03. we rewrite this as 40,000 times 1.03. This 03:58:15.659 --> 03:58:26.119 is your original salary multiplied by a growth factor of 1.03. After two years, you'll get 03:58:26.119 --> 03:58:33.418 a 3% raise from your previous year salary, your previous year salary was 40,000 times 03:58:33.418 --> 03:58:34.418 1.03. 03:58:34.418 --> 03:58:36.668 But you'll add 03:58:36.668 --> 03:58:45.559 3% of that, again, I can think of the first number is one times 40,000 times 1.03. And 03:58:45.559 --> 03:58:53.828 I can factor out the common factor of 40,000 times 1.03 from both terms, to get 40,000 03:58:53.828 --> 03:59:08.920 times 1.03 times one plus 0.03. Let me rewrite that as 40,000 times 1.03 times 1.03 or 40,000 03:59:08.920 --> 03:59:19.068 times 1.3 squared. We can think of this as your last year salary multiplied by the same 03:59:19.068 --> 03:59:26.799 growth factor of 1.03. After three years, a similar computation will give you that your 03:59:26.799 --> 03:59:37.019 new salary is your previous year salary times that growth factor of 1.03 that can be written 03:59:37.020 --> 03:59:45.720 as 40,000 times 1.03 cubed. And in general, if you're noticing the pattern, after two 03:59:45.719 --> 03:59:53.760 years, your salary should be 40,000 times 1.03 to the t power. In other words, your 03:59:53.760 --> 04:00:06.260 salary after two years is your original salary multiplied By the growth factor of 1.03, taken 04:00:06.260 --> 04:00:15.818 to the t power, let me write this as a formula s of t, where S of t is your salary is equal 04:00:15.818 --> 04:00:24.969 to 40,000 times 1.03 to the T. This is an exponential function, that is a function of 04:00:24.969 --> 04:00:38.760 the form a times b to the T, where your initial value a is 40,000 and your base b is 1.03. 04:00:38.760 --> 04:00:45.050 Notice that your base is the amount that your salary gets multiplied by each year. Given 04:00:45.050 --> 04:00:49.868 this formula, we can easily figure out what your salary will be after, for example, five 04:00:49.869 --> 04:01:01.279 years by plugging in five for T. I worked that out on my calculator to be $46,370.96 04:01:01.279 --> 04:01:08.459 to the nearest cent. exponential functions are also useful in modeling population growth. 04:01:08.459 --> 04:01:15.209 The United Nations estimated that the world population in 2010 was 6.7 9 billion growing 04:01:15.209 --> 04:01:20.739 at a rate of 1.1% per year. Assuming that the growth rate stayed the same and will continue 04:01:20.738 --> 04:01:26.309 to stay the same, we'll write an equation for the population at t years after the year 04:01:26.309 --> 04:01:39.109 2010 1.1%, written as a decimal is 0.011. So if we work out a chart as before, we see 04:01:39.109 --> 04:01:46.998 that after zero years since 2010, we have our initial population 6.7 9 billion, after 04:01:46.998 --> 04:01:59.680 one year, we'll take that 6.7 9 billion and add 1.1% of it. That is point 011 times 6.79. 04:01:59.680 --> 04:02:15.898 This works out to 6.79 times one plus point 011 or 6.79 times 1.011. Here we have our 04:02:15.898 --> 04:02:27.599 initial population of 6.7 9 billion, and our growth factor of 1.011. That's how much the 04:02:27.600 --> 04:02:35.640 population got multiplied by in one year. As before, we can work out that after two 04:02:35.639 --> 04:02:45.010 years, our population becomes 6.79 times 1.011 squared, since they got multiplied by 1.011, 04:02:45.010 --> 04:02:56.488 twice, and after two years, it'll be 6.79 times 1.011 to the t power. So our function 04:02:56.488 --> 04:03:09.520 that models population is going to be 6.79 times 1.011 to the T. Here, t represents time 04:03:09.520 --> 04:03:21.220 in years, since 2010. Just for fun, I'll plug in t equals 40. That's 40 years since 2010. 04:03:21.219 --> 04:03:31.408 So that's the year 2050. And I get 6.79 times 1.011 to the 40th power, which works out to 04:03:31.408 --> 04:03:40.368 10 point 5 billion. That's the prediction based on this exponential model. 04:03:40.369 --> 04:03:45.629 The previous two examples were examples of exponential growth. This last example is an 04:03:45.629 --> 04:03:51.869 example of exponential decay. The drugs Seroquel is metabolized and eliminated from the body 04:03:51.869 --> 04:03:58.959 at a rate of 11% per hour. If 400 milligrams are given how much remains in the body. 24 04:03:58.959 --> 04:04:09.090 hours later, I'll chart out my information where the left column will be time in hours 04:04:09.090 --> 04:04:15.998 since the dose was given, and the right column will be the number of milligrams of Seroquel 04:04:15.998 --> 04:04:22.329 still on the body. zero hours after the dose is given, we have the full 400 milligrams 04:04:22.329 --> 04:04:30.668 in the body. One hour later, we have the formula 100 milligrams minus 11% of it, that's minus 04:04:30.668 --> 04:04:40.078 point one one times 400. If I factor out the 400 from both terms, I get 400 times one minus 04:04:40.078 --> 04:04:53.219 0.11 or 400 times point eight nine. The 400 represents the initial amount. The point eight 04:04:53.219 --> 04:04:58.608 nine I'll call the growth factor, even though in this case the quantity is decreasing, not 04:04:58.609 --> 04:05:08.440 growing. So really it's kind of a shrink factor. Let's see what happens after two hours. Now 04:05:08.440 --> 04:05:15.550 I'll have 400 times 0.89 my previous amount, it'll again get multiplied by point eight, 04:05:15.549 --> 04:05:22.469 nine, so that's going to be 400 times 0.89 squared. And in general, after t hours, I'll 04:05:22.469 --> 04:05:29.528 have 400 multiplied by this growth or shrinkage factor of point eight nine, raised to the 04:05:29.529 --> 04:05:36.270 t power. Since each hour, the amount of Seroquel gets multiplied by point eight, nine, that 04:05:36.270 --> 04:05:44.289 number less than one. All right, my exponential decay function as f of t equals 400 times 04:05:44.289 --> 04:05:54.560 0.89 to the T, where f of t represents the number of milligrams of Seroquel in the body. 04:05:54.559 --> 04:06:01.958 And t represents the number of hours since the dose was given. To find out how much is 04:06:01.959 --> 04:06:13.239 in the body after 24 hours, I just plug in 24 for t. This works out to about 24.4 milligrams, 04:06:13.238 --> 04:06:17.788 I hope you notice the common form for the functions used to model all three of these 04:06:17.789 --> 04:06:26.418 examples. The functions are always in the form f of t equals a times b to the T, where 04:06:26.418 --> 04:06:36.449 a represented the initial amount, and B represented the growth factor. To find the growth factor 04:06:36.449 --> 04:06:45.449 B, we started with the percent increase or decrease. We wrote it as a decimal. And then 04:06:45.449 --> 04:06:50.800 we either added or subtracted it from one, depending on whether the quantity was increasing 04:06:50.799 --> 04:06:58.738 or decreasing. Let me show you that as a couple examples. In the first example, we had a percent 04:06:58.738 --> 04:07:09.010 increase of 3% on the race as a decimal, I'll call this r, this was point O three. And to 04:07:09.010 --> 04:07:16.619 get the growth factor, we added that to one to get 1.03. In the world population example, 04:07:16.619 --> 04:07:29.640 we had a 1.1% increase, we wrote this as point 011 and added it to one to get 1.011. In the 04:07:29.639 --> 04:07:42.349 drug example, we had a decrease of 11%. We wrote that as a decimal. And we subtracted 04:07:42.350 --> 04:07:50.988 the point one one from one to get 0.89. In fact, you can always write the growth factor 04:07:50.988 --> 04:07:58.838 B as one plus the percent change written as a decimal. If you're careful to make that 04:07:58.838 --> 04:08:04.539 percent change, negative when the quantity is decreasing and positive when the quantity 04:08:04.539 --> 04:08:09.189 is increasing. Since here one plus negative point 11 04:08:09.189 --> 04:08:14.210 gives us the correct growth factor of point eight nine which is less than one as a good 04:08:14.209 --> 04:08:23.599 check. Remember that if your quantity is increasing, the base b should be bigger than one. And 04:08:23.600 --> 04:08:30.529 if the quantity is decreasing, then B should be less than one. 04:08:30.529 --> 04:08:35.129 exponential functions can also be used to model bank accounts and loans with compound 04:08:35.129 --> 04:08:42.229 interest, as we'll see in another video. This video is about interpreting exponential functions. 04:08:42.228 --> 04:08:49.728 An antique car is worth $50,000 now, and its value increases by 7%. Each year, let's write 04:08:49.728 --> 04:09:00.448 an equation to model its value x years from now. After one year, it's value is the 50,000 04:09:00.449 --> 04:09:10.459 plus 0.07 times the 50,000. That's because its value has grown by 7% or point oh seven 04:09:10.459 --> 04:09:20.199 times 50,000. this can be written as 50,000 times one plus point O seven. Notice that 04:09:20.199 --> 04:09:28.359 adding 7% to the original value is the same as multiplying the original value by one plus 04:09:28.359 --> 04:09:41.770 point oh seven or by 1.07. After two years, the value will be 50,000 times one plus 0.07 04:09:41.770 --> 04:09:51.748 squared, or 50,000 times 1.07 squared. That's because the previous year's value is multiplied 04:09:51.748 --> 04:10:06.408 again by 1.7. In general, after x years We have x, the antique cars value will be 50,000 04:10:06.408 --> 04:10:16.728 times 1.07 to the x. That's because the original value of 50,000 gets multiplied by 1.07x times 04:10:16.728 --> 04:10:24.019 one time for each year. If we dissect this equation, we see that the number of 50,000 04:10:24.020 --> 04:10:29.709 comes from the original value of the car. The one point is seven, which I call the growth 04:10:29.709 --> 04:10:41.949 factor comes from one plus point 707. The point O seven being the, the percent increase 04:10:41.949 --> 04:10:50.010 written as a decimal. So the form of this equation is an exponential equation, V of 04:10:50.010 --> 04:10:57.738 x equals a times b to the x, where A is the initial value, and B is the growth factor. 04:10:57.738 --> 04:11:06.478 But we could also write this as a times one plus r to the x, where A is still the initial 04:11:06.478 --> 04:11:07.478 value, 04:11:07.478 --> 04:11:08.478 but 04:11:08.478 --> 04:11:15.709 R is the percent increase written as a decimal. This same equation will come up in the next 04:11:15.709 --> 04:11:22.559 example, here, my Toyota Prius is worth only 3000. Now, and its value is decreasing by 04:11:22.559 --> 04:11:34.889 5% each year. So after one year, its value will be 3000 minus 0.05 times 3000. That is 04:11:34.889 --> 04:11:48.738 3000 times one minus 0.05. I can also write that as 3000 times 0.95. Decreasing the value 04:11:48.738 --> 04:12:00.020 by 5% is like multiplying the value by one minus point oh five, or by point nine, five. 04:12:00.020 --> 04:12:08.658 After two years, the value will be multiplied by point nine five again. So the value will 04:12:08.658 --> 04:12:17.908 be 3000 times point nine, five squared. And after x years, the value will be 3000 times 04:12:17.908 --> 04:12:26.998 point nine five to the x. So my equation for the value is 3000 times point nine five to 04:12:26.998 --> 04:12:36.408 the x. This is again, an equation of the form V of x equals a times b to the x, where here, 04:12:36.408 --> 04:12:47.588 a is 3000, the initial value, and B is point nine, five, I'll still call that the growth 04:12:47.588 --> 04:12:53.738 factor, even though we're actually declining value not growing. Now remember where this 04:12:53.738 --> 04:13:01.788 point nine five came from, it came from taking one and subtracting point 05, because of the 04:13:01.789 --> 04:13:09.129 5%, decrease in value, so I can again, write my equation in the form a times well, this 04:13:09.129 --> 04:13:19.829 time times one minus r to the x, where R is our point 05. That's our percent decrease 04:13:19.828 --> 04:13:26.920 written as a decimal. Please take a moment to study this equation and the previous one. 04:13:26.920 --> 04:13:35.408 They say that when you have an exponential function, the number here is the initial value. 04:13:35.408 --> 04:13:42.088 If it's written in this form, B is your growth factor. But you can think of B as being either 04:13:42.088 --> 04:13:50.510 one minus r, where r is the percent decrease, or one plus r, where r is the percent increase. 04:13:50.510 --> 04:13:55.898 In this example, we're given a function f of x to model the number of bacteria in a 04:13:55.898 --> 04:14:03.510 petri dish x hours after 12 o'clock noon, we want to know what was the number of bacteria 04:14:03.510 --> 04:14:09.168 at noon, and by what percent, the number of bacteria is increasing every hour, we can 04:14:09.168 --> 04:14:14.408 see from the equation that the number of bacteria is increasing and not decreasing, because 04:14:14.408 --> 04:14:21.458 the base of the exponential function 1.45 is bigger than one. Notice that our equation 04:14:21.459 --> 04:14:30.189 f of x equals 12 times 1.45 to the X has the form of a times b to the x, or we can think 04:14:30.189 --> 04:14:43.389 of it as a times one plus r to the x. Here a is 12, b is 1.45 and r is 0.45. Based on 04:14:43.389 --> 04:14:49.158 this familiar form, we can recognize that the initial amount of bacteria is going to 04:14:49.158 --> 04:15:00.949 be 12 12,000. Since those are our units and this value 1.45 is our growth factor. What 04:15:00.949 --> 04:15:08.859 the number of bacteria is multiplied by each hour? Well are the point four five is the 04:15:08.859 --> 04:15:18.529 the rate of increase, in other words, a 45% increase each hour. So our answers to the 04:15:18.529 --> 04:15:29.359 questions are 12,045%. In this example, the population of salamanders is modeled by this 04:15:29.359 --> 04:15:37.529 exponential function, where x is the number of years since 2015. Notice that the number 04:15:37.529 --> 04:15:42.850 of salamanders is decreasing, because the base of our exponential function point seven, 04:15:42.850 --> 04:15:50.100 eight is less than one. So if we recognize the form of our exponential function, a times 04:15:50.100 --> 04:15:59.579 b to the x, or we can think of this as a times one minus r to the x, where A is our initial 04:15:59.579 --> 04:16:08.998 value, and r is our percent decrease written as a decimal. 04:16:08.998 --> 04:16:18.779 Our initial value is 3000. So that's the number of salamanders zero years after 2015. Our 04:16:18.779 --> 04:16:27.529 growth factor B is 0.78. But if I write that as one minus r, I see that R has to be one 04:16:27.529 --> 04:16:40.720 minus 0.78, or 0.22. In other words, our population is decreasing by 22% each year. In this video, 04:16:40.719 --> 04:16:47.028 we saw that exponential functions can be written in the form f of x equals a times b to the 04:16:47.029 --> 04:16:58.050 x, where A is the initial value. And B is the growth factor. We also saw that they can 04:16:58.049 --> 04:17:05.878 be written in the form a times one plus r to the x when the amount is increasing. And 04:17:05.879 --> 04:17:16.609 as a times one minus r to the x when the amount is decreasing. In this format, R is the percent 04:17:16.609 --> 04:17:28.350 increase or the percent decrease written as a decimal. So 15% increase will be an R value 04:17:28.350 --> 04:17:43.690 of 0.15 and a growth factor B of 1.15. Whereas a 12% decrease will be an R value of point 04:17:43.690 --> 04:17:55.779 one, two, and a B value of one minus point one, two, or 0.7. Sorry, 0.88. These observations 04:17:55.779 --> 04:18:03.439 help us quickly interpret exponential functions. For example, here, we have an initial value 04:18:03.439 --> 04:18:17.889 of 100 and a 15% increase. And here, we have an initial value of 50 and a 40%. decrease. 04:18:17.889 --> 04:18:23.118 exponential functions can be used to model compound interest for loans and bank accounts. 04:18:23.119 --> 04:18:30.291 Suppose you invest $200 in a bank account that earns 3% interest every year. If you 04:18:30.290 --> 04:18:37.418 make no deposits or withdrawals, how much money will you have accumulated after 10 years, 04:18:37.418 --> 04:18:41.939 because 3% of the money that's in the bank is getting added each year, the money in the 04:18:41.939 --> 04:18:51.359 bank gets multiplied by 1.03 each year. So after one year, the amount will be 200 times 04:18:51.359 --> 04:19:02.180 1.3, after two years 200 times one point O three squared, and and after t years 200 times 04:19:02.180 --> 04:19:08.479 one point O three to the t power. So the function modeling the amount of money, I'll call it 04:19:08.478 --> 04:19:20.418 P of t is given by 200 times 1.03 to the T. More generally, if you invest a dollars 04:19:20.418 --> 04:19:32.628 at an annual interest rate of our for t years. In the end, you'll have P of t equals a times 04:19:32.629 --> 04:19:37.079 one plus r to the T 04:19:37.078 --> 04:19:45.039 here r needs to be written as a decimal, so 0.03 In our example, for the 3% annual interest 04:19:45.040 --> 04:19:51.909 rate. Going back to our specific example, After 10 years, the amount of money is going 04:19:51.908 --> 04:20:03.038 to be P of 10 which is 200 times 1.03 to the 10 which works out to 206 $68.78 to the nearest 04:20:03.039 --> 04:20:10.239 cent. In this problem, we've assumed that the interest accumulates once per year. But 04:20:10.238 --> 04:20:14.510 in the next few examples, we'll see what happens when the interest rate accumulates more frequently, 04:20:14.510 --> 04:20:23.079 twice a year, or every month. For example, let's deposit $300 in an account that earns 04:20:23.079 --> 04:20:31.059 4.5% annual interest compounded semi annually, this means two times a year or every six months. 04:20:31.059 --> 04:20:40.748 A 4.5% annual interest rate compounded two times a year means that we're actually getting 04:20:40.748 --> 04:20:53.039 4.5 over 2% interest, every time the interest is compounded. That is every six months. That's 04:20:53.040 --> 04:21:05.890 2.25% interest every half a year. Note that 2.25% is the same as 0.02 to five as a decimal. 04:21:05.889 --> 04:21:17.128 So every time we earn interest, our money gets multiplied by 1.02 to five. Let me make 04:21:17.129 --> 04:21:25.399 a chart of what happens. After zero years, which is also zero half years, we have our 04:21:25.398 --> 04:21:33.760 original $300. for half a year, that's one half year, our money gets earns interest one 04:21:33.760 --> 04:21:42.078 time, so we multiply the 300 by 1.02 to five, after one year, that's two half years, our 04:21:42.078 --> 04:21:52.770 money earns interest two times. So we multiply 300 by 1.0225 squared. continuing this way, 04:21:52.770 --> 04:22:01.680 after 1.5 years, that's three half years, we have 300 times 1.02 to five cubed. And 04:22:01.680 --> 04:22:09.750 after two years or four half years, we have 300 times 1.02 to five to the fourth. In general, 04:22:09.750 --> 04:22:20.158 after t years, which is to T half years, our money will grow to 300 times 1.02 to five 04:22:20.158 --> 04:22:28.299 to the two t power. Because we've compounded interest to tee times our formula for our 04:22:28.299 --> 04:22:35.328 amount of money is P of t equals 300 times one point O two to five to the two t where 04:22:35.328 --> 04:22:45.639 t is the number of years. To finish the problem, after seven years, we'll have P of $7, which 04:22:45.639 --> 04:22:59.849 is 300 times 1.02 to five to the two times seven or 14 power. And that works out to $409.65 04:22:59.850 --> 04:23:06.470 to the nearest cent. In this next example, we're going to take out a loan for $1,200 04:23:06.469 --> 04:23:14.719 in annual interest rate of 6% compounded monthly. Although loans and bank accounts might feel 04:23:14.719 --> 04:23:19.130 different, they're mathematically the same. It's like from the bank's point of view, they're 04:23:19.130 --> 04:23:25.278 investing money in you and getting interest on their money from you. So we can work them 04:23:25.279 --> 04:23:36.050 out with the same kind of math 6% annual interest rate compounded monthly means you're compounding 04:23:36.049 --> 04:23:42.250 12 times a year. So each time you compound interest, you're just going to get six over 04:23:42.250 --> 04:23:56.148 12% interest. That's point 5% interest. And as a decimal, that's 0.005. Let's try it out 04:23:56.148 --> 04:24:05.689 again, what happens. Time is zero, of course, you'll have the original loan amount of 1200. 04:24:05.689 --> 04:24:15.809 After one year, that's 12 months, your loan has had interest added to it 12 times so it 04:24:15.809 --> 04:24:22.510 gets multiplied by 1.05 to the 12th. 04:24:22.510 --> 04:24:30.648 After two years, that's 24 months, it's had interest added to it 24 times, so it gets 04:24:30.648 --> 04:24:39.108 multiplied by 1.05 to the 24th power. Similarly, after three years or 36 months, your loan 04:24:39.109 --> 04:24:49.890 amount will be 1200 times 1.05 to the 36th power. And in general, after two years, that's 04:24:49.889 --> 04:24:59.939 12 t months. So the interest will be compounded 12 t times and so we have to raise the 1.05 04:24:59.939 --> 04:25:08.120 To the 12 t power. This gives us the general formula for the money owed is P of t equals 04:25:08.120 --> 04:25:20.399 1200 times 1.05 to the 12 T, where T is the number of years. In particular, after three 04:25:20.398 --> 04:25:31.398 years, we'll have to pay back a total of 1200 times 1.05 to the 12 times three or 36 power, 04:25:31.398 --> 04:25:42.799 which works out to $1,436.02 to the nearest cent. These last two problems follow a general 04:25:42.799 --> 04:25:51.938 pattern. If A is the initial amount of the loan or bank account, and r is the annual 04:25:51.939 --> 04:26:03.529 interest rate, compounded n times per year, then our formula for the amount of money is 04:26:03.529 --> 04:26:13.620 going to be a times one plus r over n to the n t. This formula exactly matches what we 04:26:13.620 --> 04:26:20.680 did in this problem. First, we took the interest rate our which was 6%, and divided it by the 04:26:20.680 --> 04:26:25.838 number of compounding periods each year 12. We wrote that as a decimal and added one to 04:26:25.838 --> 04:26:35.059 it. That's where we got the 1.05 from. We raised this to not to the number of years, 04:26:35.059 --> 04:26:39.818 but to 12 times the number of years. That's the number of compounding periods per year, 04:26:39.818 --> 04:26:46.628 times the number of years. And we multiplied all that by the initial amount of money, which 04:26:46.629 --> 04:26:52.979 was 1200. This formula for compound interest is a good one to memorize. But it's also important 04:26:52.978 --> 04:26:57.408 to be able to reason your way through it, like we did in this chart. There's one more 04:26:57.408 --> 04:27:02.908 type of compound interest. And that's interest compounded continuously. You can think of 04:27:02.908 --> 04:27:09.950 continuous compounding as the limit of compounding more and more frequently 10 times a year, 04:27:09.950 --> 04:27:14.969 100 times a year 1000 times a year a million times a year. In the limit, you get continuous 04:27:14.969 --> 04:27:24.809 compounding. The formula for continuous compounding is P of t equals a times e to the RT, where 04:27:24.809 --> 04:27:37.269 p of t is the amount of money, t is the time in years. A is the initial amount of money. 04:27:37.270 --> 04:27:49.120 And R is the annual interest rate written as a decimal. So 0.025 in this problem from 04:27:49.120 --> 04:27:57.000 the 2.5% annual interest rate. He represents the famous constant Oilers constant, which 04:27:57.000 --> 04:28:12.059 is about 2.718. So in this problem, we have P of t is 4000 times e to the 0.025 T. And 04:28:12.059 --> 04:28:21.549 after five years, we'll have P of five, which is 4000 times e to the 0.0 to five times five, 04:28:21.549 --> 04:28:32.228 which works out to $4,532.59 to the nearest cent. To summarize, if R represents the annual 04:28:32.228 --> 04:28:43.398 interest rate written as a decimal, that is 2% would be 0.02. And t represents the number 04:28:43.398 --> 04:28:50.719 of years and a represents the initial amount of money, then for just simple annual interest 04:28:50.719 --> 04:29:00.608 compounded once a year. Our formula is P of t is a times one plus r to the T for compound 04:29:00.609 --> 04:29:08.309 interest compounded n times per year. Our formula is P of t is a times one plus r over 04:29:08.309 --> 04:29:16.930 n to the n T. And for compound interest compounded continuously, we get P of t is a times e to 04:29:16.930 --> 04:29:18.430 the RT. 04:29:18.430 --> 04:29:24.670 In this video, we looked at three kinds of compound interest problems, simple annual 04:29:24.670 --> 04:29:33.908 interest, interest compounded and times per year, and continuously compounded interest. 04:29:33.908 --> 04:29:42.118 This video introduces logarithms. logarithms are a way of writing exponents. The expression 04:29:42.119 --> 04:29:52.949 log base a of B equals c means that a to the C equals b. In other words, log base a of 04:29:52.949 --> 04:30:02.119 B is the exponent that you raise a to to get BE THE NUMBER A It is called the base of the 04:30:02.119 --> 04:30:07.489 logarithm. It's also called the base when we write it in this exponential form. Some 04:30:07.488 --> 04:30:14.139 students find it helpful to remember this relationship. log base a of B equals c means 04:30:14.139 --> 04:30:17.778 a to the C equals b, by drawing arrows, 04:30:17.779 --> 04:30:20.510 a to the C equals b. 04:30:20.510 --> 04:30:30.779 Other students like to think of it in terms of asking a question, log base a of B, asks, 04:30:30.779 --> 04:30:41.439 What power do you raise a to in order to get b? Let's look at some examples. log base two 04:30:41.439 --> 04:30:50.408 of eight is three, because two to the three equals eight. In general, log base two of 04:30:50.408 --> 04:30:59.359 y is asking you the question, What power do you have to raise to to to get y? So for example, 04:30:59.359 --> 04:31:08.130 log base two of 16 is four, because it's asking you the question to what power equals 16? 04:31:08.129 --> 04:31:15.128 And the answer is four. Please pause the video and try some of these other examples. log 04:31:15.129 --> 04:31:22.079 base two of two is asking, What power do you raise to two to get two? And the answer is 04:31:22.079 --> 04:31:34.369 one. Two to the one equals two. log base two of one half is asking two to what power gives 04:31:34.369 --> 04:31:40.790 you one half? Well, to get one half, you need to raise two to a negative power. So that 04:31:40.790 --> 04:31:49.699 would be two to the negative one. So the answer is negative one. log base two of 1/8 means 04:31:49.699 --> 04:31:58.658 what power do we raise to two in order to get 1/8. Since 1/8, is one over two cubed, 04:31:58.658 --> 04:32:06.418 we have to raise two to the negative three power to get one over two cubed. So our exponent 04:32:06.418 --> 04:32:13.689 is negative three. And that's our answer to our log expression. Finally, log base two 04:32:13.689 --> 04:32:21.979 of one is asking to what power equals one. Well, anything raised to the zero power gives 04:32:21.978 --> 04:32:28.918 us one, so this log expression evaluates to zero. Notice that we can get positive negative 04:32:28.918 --> 04:32:36.299 and zero answers for our logarithm expressions. Please pause the video and figure out what 04:32:36.299 --> 04:32:46.929 these logs evaluate to. to work out log base 10 of a million. Notice that a million is 04:32:46.930 --> 04:32:54.078 10 to the sixth power. Now we're asking the question, What power do we raise tend to to 04:32:54.078 --> 04:32:59.748 get a million? So that is what power do we raise 10 to to get 10 to the six? Well, of 04:32:59.748 --> 04:33:11.030 course, the answer is going to be six. Similarly, since point O one is 10 to the minus three, 04:33:11.030 --> 04:33:16.271 this log expression is the same thing as asking, what's the log base 10 of 10 to the minus 04:33:16.271 --> 04:33:22.859 three? Well, what power do you have to raise 10? to to get 10 to the minus three? Of course, 04:33:22.859 --> 04:33:31.020 the answer is negative three. Log base 10 of zero is asking, What power do we raise 04:33:31.020 --> 04:33:38.189 10 to to get zero. If you think about it, there's no way to raise 10 to an exponent 04:33:38.188 --> 04:33:43.468 get zero. Raising 10 to a positive exponent gets us really big positive numbers. Raising 04:33:43.469 --> 04:33:49.840 10 to a negative exponent is like one over 10 to a power that's giving us tiny fractions, 04:33:49.840 --> 04:33:55.000 but they're still positive numbers, we're never going to get zero. Even if we raise 04:33:55.000 --> 04:33:59.919 10 to the zero power, we'll just get one. So there's no way to get zero and the log 04:33:59.919 --> 04:34:05.599 base 10 of zero does not exist. If you try it on your calculator using the log base 10 04:34:05.599 --> 04:34:11.159 button, you'll get an error message. Same thing happens when we do log base 10 of negative 04:34:11.159 --> 04:34:17.109 100. We're asking 10 to what power equals negative 100. And there's no exponent that 04:34:17.109 --> 04:34:23.170 will work. And more generally, it's possible to take the log of numbers that are greater 04:34:23.169 --> 04:34:29.929 than zero, but not for numbers that are less than or equal to zero. In other words, the 04:34:29.930 --> 04:34:36.309 domain of the function log base a of x, no matter what base you're using for a, the domain 04:34:36.309 --> 04:34:44.131 is going to be all positive numbers. A few notes on notation. When you see ln of x, that's 04:34:44.131 --> 04:34:50.729 called natural log, and it means the log base e of x where he is that famous number that's 04:34:50.729 --> 04:34:59.041 about 2.718. When you see log of x with no base at all, by convention, that means log 04:34:59.041 --> 04:35:05.770 base 10 of x And it's called the common log. Most scientific calculators have buttons for 04:35:05.770 --> 04:35:13.449 natural log, and for common log. Let's practice rewriting expressions with logs in them. log 04:35:13.449 --> 04:35:20.949 base three of one nine is negative two, can be rewritten as the expression three to the 04:35:20.949 --> 04:35:26.100 negative two equals 1/9. 04:35:26.099 --> 04:35:34.750 Log of 13 is shorthand for log base 10 of 13. So that can be rewritten as 10 to the 04:35:34.750 --> 04:35:45.760 1.11394 equals 13. Finally, in this last expression, ln means natural log, or log base e, so I 04:35:45.760 --> 04:35:52.878 can rewrite this equation as log base e of whenever E equals negative one. Well, that 04:35:52.879 --> 04:36:00.898 means the same thing as e to the negative one equals one over e, which is true. Now 04:36:00.898 --> 04:36:05.559 let's go the opposite direction. We'll start with exponential equations and rewrite them 04:36:05.559 --> 04:36:14.600 as logs. Remember that log base a of B equals c means the same thing as a to the C equals 04:36:14.599 --> 04:36:23.979 b, the base stays the same in both expressions. So for this example, the base of three in 04:36:23.979 --> 04:36:29.317 the exponential equation, that's going to be the same as the base in our log. Now I 04:36:29.317 --> 04:36:33.438 just have to figure out what's in the argument of the log. And what goes on the other side 04:36:33.438 --> 04:36:40.519 of the equal sign. Remember that the answer to a log is an exponent. So the thing that 04:36:40.520 --> 04:36:48.067 goes in this box should be my exponent for my exponential equation. In other words, you. 04:36:48.067 --> 04:36:57.919 And I'll put the 9.78 as the argument of my log. This works, because log base three of 04:36:57.919 --> 04:37:06.108 9.78 equals view means the same thing as three to the U equals 9.78, which is just what we 04:37:06.109 --> 04:37:13.831 started with. In the second example, the base of my exponential equation is E. So the base 04:37:13.830 --> 04:37:22.051 of my log is going to be the answer to my log is an exponent. In this case, the exponent 04:37:22.051 --> 04:37:32.430 3x plus seven. And the other expression, the four minus y becomes my argument of my log. 04:37:32.430 --> 04:37:41.638 Let me check, log base e of four minus y equals 3x plus seven means e to the 3x plus seven 04:37:41.638 --> 04:37:47.159 equals four minus y, which is just what I started with. I can also rewrite log base 04:37:47.159 --> 04:37:56.409 e as natural log. This video introduced the idea of logs, and the fact that log base a 04:37:56.409 --> 04:38:07.618 of B equal c means the same thing as a to the C equals b. So log base a of B is asking 04:38:07.618 --> 04:38:16.669 you the question, What power exponent Do you raise a to in order to get b. In this video, 04:38:16.669 --> 04:38:22.968 we'll work out the graph, so some log functions and also talk about their domains. For this 04:38:22.969 --> 04:38:29.240 first example, let's graph a log function by hand by plotting some points. The function 04:38:29.240 --> 04:38:36.648 we're working with is y equals log base two of x, I'll make a chart of x and y values. 04:38:36.648 --> 04:38:41.229 Since we're working this out by hand, I want to pick x values for which it's easy to compute 04:38:41.229 --> 04:38:47.879 log base two of x. So I'll start out with the x value of one because log base two of 04:38:47.879 --> 04:38:56.789 one is zero, log base anything of one is 02 is another x value that's easy to compute 04:38:56.789 --> 04:39:03.809 log base two of two, that's asking, What power do I raise to two to get to one? And the answer 04:39:03.809 --> 04:39:11.969 is one. Power other powers of two are easy to work with. So for example, log base two 04:39:11.969 --> 04:39:18.340 of four that saying what power do I raise to two to get four, so the answer is two. 04:39:18.340 --> 04:39:26.349 Similarly, log base two of eight is three and log base two of 16 is four. Let me also 04:39:26.349 --> 04:39:33.919 work with some fractional values for X. If x is one half, then log base two of one half 04:39:33.919 --> 04:39:39.329 that saying what power do I raise to two to get one half? Well, that needs a power of 04:39:39.330 --> 04:39:48.430 negative one. It's also easy to compute by hand, the log base two of 1/4 and 1/8. log 04:39:48.430 --> 04:39:57.740 base two of 1/4 is negative two since two to the negative two is 1/4. And similarly, 04:39:57.740 --> 04:40:04.850 log base two of one eight is negative f3 I'll put some tick marks on my x and y axes. Please 04:40:04.849 --> 04:40:10.817 pause the video and take a moment to plot these points. Let's see, I have the point, 04:40:10.817 --> 04:40:17.579 one zero, that's here to one that's here, for two, 04:40:17.580 --> 04:40:26.190 that is here, and then eight, three, which is here. And the fractional x values, one 04:40:26.189 --> 04:40:33.457 half goes with negative one, and 1/4 with negative two 1/8 with negative three. And 04:40:33.457 --> 04:40:39.739 if I connect the dots, I get a graph that looks something like this. If I had smaller 04:40:39.740 --> 04:40:44.700 and smaller fractions, I would keep getting more and more negative answers when I took 04:40:44.700 --> 04:40:51.200 log base two of them, so my graph is getting more and more negative, my y values are getting 04:40:51.200 --> 04:40:56.920 more and more negative, as x is getting close to zero. Now I didn't draw any parts of the 04:40:56.919 --> 04:41:02.189 graph over here with negative X values, I didn't put any negative X values in my chart, 04:41:02.189 --> 04:41:08.599 that omission is no accident. Because if you try to take the log base two or base anything 04:41:08.599 --> 04:41:15.829 of a negative number, like say negative four or something, there's no answer. This doesn't 04:41:15.830 --> 04:41:24.951 exist because there's no power that you can raise to to to get a negative number. So there 04:41:24.951 --> 04:41:29.360 are no points on the graph for negative X values. And similarly, there are no points 04:41:29.360 --> 04:41:35.549 on the graph where x is zero, because you can't take log base two of zero, there's no 04:41:35.549 --> 04:41:42.378 power you can raise to two to get zero. I want to observe some key features of this 04:41:42.378 --> 04:41:51.297 graph. First of all, the domain is x values greater than zero. In interval notation, I 04:41:51.297 --> 04:41:58.029 can write that as a round bracket because I don't want to include zero to infinity, 04:41:58.029 --> 04:42:05.039 the range is going to be the y values, while they go all the way down into the far reaches 04:42:05.040 --> 04:42:11.069 of the negative numbers. And the graph gradually increases y value is getting bigger and bigger. 04:42:11.069 --> 04:42:17.279 So the range is actually all real numbers are an interval notation negative infinity 04:42:17.279 --> 04:42:25.590 to infinity. Finally, I want to point out that this graph has a vertical asymptote at 04:42:25.590 --> 04:42:35.887 the y axis, that is at the line x equals zero. I'll draw that on my graph with a dotted line. 04:42:35.887 --> 04:42:42.569 A vertical asymptote is a line that our functions graph gets closer and closer to. So this is 04:42:42.569 --> 04:42:50.189 the graph of y equals log base two of x. But if I wanted to graph say, y equals log base 04:42:50.189 --> 04:42:57.250 10 of x, it would look very similar, it would still have a domain of X values greater than 04:42:57.250 --> 04:43:02.878 zero, a range of all real numbers and a vertical asymptote at the y axis, it will still go 04:43:02.878 --> 04:43:10.360 through the point one zero, but it would go through the point 10 one instead, because 04:43:10.360 --> 04:43:20.159 log base 10 of 10 is one, it would look pretty much the same, just a lot flatter over here. 04:43:20.159 --> 04:43:24.930 But even though it doesn't look like it with the way I've drawn it, it's still gradually 04:43:24.930 --> 04:43:34.950 goes up to n towards infinity. In fact, the graph of y equals log base neaa of X for a 04:43:34.950 --> 04:43:42.387 bigger than one looks pretty much the same, and has the same three properties. Now that 04:43:42.387 --> 04:43:46.969 we know what the basic log graph looks like, we can plot at least rough graphs of other 04:43:46.970 --> 04:43:52.968 log functions without plotting points. Here we have the graph of natural log of X plus 04:43:52.968 --> 04:43:57.750 five. And again, I'm just going to draw a rough graph. If I did want to do a more accurate 04:43:57.750 --> 04:44:02.477 graph, I probably would want to plot some points. But I know that roughly a log graph, 04:44:02.477 --> 04:44:09.930 if it was just like y equals ln of x, that would look something like this, and it would 04:44:09.930 --> 04:44:17.271 go through the point one zero, with a vertical asymptote along the y axis. Now if I want 04:44:17.271 --> 04:44:23.389 a graph, ln of x plus five, that just shifts our graph by five units, it'll still have 04:44:23.389 --> 04:44:27.450 the same vertical asymptote. Since the vertical line shifted up by five units, there's still 04:44:27.450 --> 04:44:33.119 a vertical line, but instead of going through one zero, it'll go through the point one, 04:44:33.119 --> 04:44:42.680 five. So I'll draw a rough sketch here. Let's compare our starting function y equals ln 04:44:42.680 --> 04:44:51.080 x and the transformed version y equals ln x plus five in terms of the domain, the range 04:44:51.080 --> 04:44:59.750 and the vertical asymptote. Our original function y equals ln x had a domain of zero to infinity 04:44:59.750 --> 04:45:08.659 Since adding five on the outside affects the y values, and the domain is the x values, 04:45:08.659 --> 04:45:16.340 this transformation doesn't change the domain. So the domain is still from zero to infinity. 04:45:16.340 --> 04:45:22.529 Now the range of our original y equals ln x was from negative infinity to infinity. 04:45:22.529 --> 04:45:28.680 Shifting up by five does affect the y values, and the range is talking about the y values. 04:45:28.680 --> 04:45:33.260 But since the original range was all real numbers, if you add five to all set of all 04:45:33.259 --> 04:45:38.147 real numbers, you still get the set of all real numbers. So in this case, the range doesn't 04:45:38.148 --> 04:45:43.958 change either. And finally, we already saw that the original vertical asymptote of the 04:45:43.957 --> 04:45:49.647 y axis x equals zero, when we shift that up by five units, it's still the vertical line 04:45:49.648 --> 04:45:57.490 x equals zero. In this next example, we're starting with a log base 10 function. And 04:45:57.490 --> 04:46:02.978 since the plus two is on the inside, that means we shift that graph left by two. So 04:46:02.977 --> 04:46:09.718 I'll draw our basic log function. Here's our basic log function. So I'll think of that 04:46:09.718 --> 04:46:15.317 as y equals log of x going through the point one, zero, 04:46:15.317 --> 04:46:23.000 here's its vertical asymptote. Now I need to shift everything left by two. So my vertical 04:46:23.000 --> 04:46:28.610 asymptote shifts left, and now it's at the line x equals negative two, instead of at 04:46:28.610 --> 04:46:37.090 x equals zero, and my graph, let's see my point, one zero gets shifted to, let's see 04:46:37.090 --> 04:46:45.270 negative one zero, since I'm subtracting two from the axis, and here's a rough sketch of 04:46:45.270 --> 04:46:54.069 the resulting graph. Let's compare the features of the two graphs drawn here. We're talking 04:46:54.069 --> 04:47:01.207 about domains, the original had a domain of from zero to infinity. But now I've shifted 04:47:01.207 --> 04:47:07.119 that left. So I've subtracted two from all my x values. And here's my new domain, which 04:47:07.119 --> 04:47:15.227 I can also verify just by looking at the picture. My range was originally from negative infinity 04:47:15.227 --> 04:47:20.218 to infinity, well, shifting left only affects the x value. So it doesn't even affect the 04:47:20.218 --> 04:47:26.690 range. So my range is still negative infinity to infinity. My vertical asymptote was originally 04:47:26.689 --> 04:47:32.967 at x equals zero. And since I subtract two from all x values, that shifts it to x equals 04:47:32.968 --> 04:47:38.190 negative two. In this last problem, I'm not going to worry about drawing this graph. I'll 04:47:38.189 --> 04:47:44.567 just use algebra to compute its domain. So let's think about what's the issue, when you're 04:47:44.567 --> 04:47:50.079 taking the logs of things? Well, you can't take the log of a negative number is zero. 04:47:50.080 --> 04:47:56.010 So whatever's inside the argument of the log function, whatever is being fed into log had 04:47:56.009 --> 04:48:03.269 better be greater than zero. So I'll write that down, we need to minus 3x to be greater 04:48:03.270 --> 04:48:08.797 than zero. Now it's a matter of solving an inequality to it's got to be greater than 04:48:08.797 --> 04:48:15.759 3x. So two thirds is greater than x. In other words, x has to be less than two thirds. So 04:48:15.759 --> 04:48:23.340 our domain is all the x values from negative infinity to two thirds, not including two 04:48:23.340 --> 04:48:31.420 thirds. It's a good idea to memorize the basic shape of the graph of a log function. It looks 04:48:31.419 --> 04:48:36.759 something like this, go through the point one zero, and has a vertical asymptote on 04:48:36.759 --> 04:48:44.798 the y axis. Also, if you remember that you can't take the log of a negative number, or 04:48:44.798 --> 04:48:52.878 zero, then that helps you quickly compute domains for log functions. Whatever's inside 04:48:52.878 --> 04:49:01.779 the log function, you set that greater than zero, and solve. This video is about combining 04:49:01.779 --> 04:49:07.500 logs and exponents. Please pause the video and take a moment to use your calculator to 04:49:07.500 --> 04:49:15.707 evaluate the following four expressions. Remember, that log base 10 on your calculator is the 04:49:15.707 --> 04:49:24.817 log button. While log base e on your calculator is the natural log button, you should find 04:49:24.817 --> 04:49:34.409 that the log base 10 of 10 cubed is three. The log base e of e to the 4.2 is 4.2 10 to 04:49:34.409 --> 04:49:45.387 the log base 10 of 1000 is 1000. And eat the log base e of 9.6 is 9.6. In each case, the 04:49:45.387 --> 04:49:51.718 log and the exponential function with the same base undo each other and we're left with 04:49:51.718 --> 04:49:59.420 the exponent. In fact, it's true that for any base a the log base a of a to the x is 04:49:59.419 --> 04:50:05.289 equal to x, the same sort of cancellation happens if we do the exponential function 04:50:05.290 --> 04:50:10.770 in the log function with the same base in the opposite order. For example, we take 10 04:50:10.770 --> 04:50:15.869 to the power of log base 10 of 1000, the 10 to the power and the log base 10 undo each 04:50:15.869 --> 04:50:21.770 other, and we're left with the 1000s. This happens for any base, say, a data log base 04:50:21.770 --> 04:50:31.067 a of x is equal to x. We can describe this by saying that an exponential function and 04:50:31.067 --> 04:50:40.169 a log function with the same base undo each other. If you're familiar with the language 04:50:40.169 --> 04:50:46.869 of inverse functions, the exponential function and log function are inverses. Let's see why 04:50:46.869 --> 04:50:55.289 these roles hold for the first log role. log base a of a dx is asking the question, What 04:50:55.290 --> 04:51:03.270 power do we raise a two in order to get a to the x? In other words, a two what power 04:51:03.270 --> 04:51:11.430 is a dx? Well, the answer is clearly x. And that's why log base a of a to the x equals 04:51:11.430 --> 04:51:15.290 x. For the second log rule, 04:51:15.290 --> 04:51:25.120 notice that the log base a of x means the power we raise a two to get x. But this expression 04:51:25.119 --> 04:51:30.137 is saying that we're supposed to raise a to that power. If we raise a to the power, we 04:51:30.137 --> 04:51:37.128 need to raise a two to get x, then we'll certainly get x. Now let's use these two rules. In some 04:51:37.128 --> 04:51:44.610 examples. If we want to find three to the log base three of 1.43 to the power and log 04:51:44.610 --> 04:51:51.860 base three undo each other, so we're left with 1.4. If we want to find ln of e iliacs. 04:51:51.860 --> 04:51:59.060 Remember that ln means log base e. So we're taking log base e of e to the x, well, those 04:51:59.060 --> 04:52:05.398 functions undo each other, and we're left with x. If we want to take 10 to the log of 04:52:05.398 --> 04:52:12.600 three z, remember that log without a base written implies that the base is 10. So really, 04:52:12.599 --> 04:52:18.750 we want to take 10 to the log base 10 of three z will tend to a power and log base 10 undo 04:52:18.750 --> 04:52:26.409 each other. So we're left with a three z. Finally, this last statement hold is ln of 04:52:26.409 --> 04:52:36.419 10 to the x equal to x, well, ln means log base e. So we're taking log base e of 10 to 04:52:36.419 --> 04:52:41.159 the x, notice that the base of the log and the base of the exponential function are not 04:52:41.159 --> 04:52:47.939 the same. So they don't undo each other. And in fact, log base e of 10 to the x is not 04:52:47.939 --> 04:52:53.919 usually equal to x, we can check with one example. Say if x equals one, then log base 04:52:53.919 --> 04:52:59.629 e of 10 to the one, that's log base e of 10. And we can check on the calculator that's 04:52:59.630 --> 04:53:06.558 equal to 2.3. And some more decimals, which is not the same thing as one. So this statement 04:53:06.558 --> 04:53:13.900 is false, it does not hold. We need the basis to be the same for logs and exponents to undo 04:53:13.900 --> 04:53:23.000 each other. In this video, we saw that logs and exponents with the same base undo each 04:53:23.000 --> 04:53:26.250 other. Specifically, 04:53:26.250 --> 04:53:33.779 log base a of a to the x is equal to x and a to the log base a of x is also equal to 04:53:33.779 --> 04:53:34.779 x 04:53:34.779 --> 04:53:42.860 for any values of x and any base a. This video is about rules or properties of logs. The 04:53:42.860 --> 04:53:46.779 log rules are closely related to the exponent rules. So let's start by reviewing some of 04:53:46.779 --> 04:53:51.378 the exponent rules. To keep things simple, we'll write everything down with a base of 04:53:51.378 --> 04:53:57.297 two. Even though the exponent rules hold for any base. We know that if we raise two to 04:53:57.297 --> 04:54:04.039 the zero power, we get one, we have a product rule for exponents, which says that two to 04:54:04.040 --> 04:54:12.120 the M times two to the n is equal to two to the m plus n. In other words, if we multiply 04:54:12.119 --> 04:54:18.590 two numbers, then we add the exponents. We also have a quotient rule that says that two 04:54:18.590 --> 04:54:26.590 to the M divided by n to the n is equal to two to the m minus n. In words, that says 04:54:26.590 --> 04:54:34.520 that if we divide two numbers, then we subtract the exponents. Finally, we have a power rule 04:54:34.520 --> 04:54:43.317 that says if we take a power to a power, then we multiply the exponents. Each of these exponent 04:54:43.317 --> 04:54:50.270 rules can be rewritten as a log rule. The first rule, two to the zero equals one can 04:54:50.270 --> 04:54:57.990 be rewritten in terms of logs as log base two of one equals zero. That's because log 04:54:57.990 --> 04:55:05.090 base two of one equals zero mean To the zero equals one. The second rule, the product rule 04:55:05.090 --> 04:55:14.990 can be rewritten in terms of logs by saying log of x times y equals log of x plus log 04:55:14.990 --> 04:55:21.990 of y. I'll make these base two to agree with my base that I'm using for my exponent rules. 04:55:21.990 --> 04:55:28.860 In words, that says the log of the product is the sum of the logs. Since logs really 04:55:28.860 --> 04:55:34.637 represent exponent, this is saying that when you multiply two numbers together, you add 04:55:34.637 --> 04:55:41.329 their exponents, which is just what we said for the exponent version. The quotient rule 04:55:41.330 --> 04:55:48.830 for exponents can be rewritten in terms of logs by saying the log of x divided by y is 04:55:48.830 --> 04:55:58.160 equal to the log of x minus the log of y. In words, we can say that the log of the quotient 04:55:58.159 --> 04:56:04.968 is equal to the difference of the logs. Since logs are really exponents, another way of 04:56:04.968 --> 04:56:11.770 saying the same thing is that when you divide two numbers, you subtract their exponents. 04:56:11.770 --> 04:56:18.600 That's how we described the exponent rule above. Finally, the power rule for exponents 04:56:18.599 --> 04:56:27.317 can be rewritten in terms of logs by saying the log of x to the n is equal to n times 04:56:27.317 --> 04:56:34.090 log of x. Sometimes people describe this rule by saying when you take the log of an expression 04:56:34.090 --> 04:56:40.148 with an exponent, you can bring down the exponent and multiply. If we think of x as being some 04:56:40.148 --> 04:56:45.409 power of two, this is really saying when we take a power to a power, we multiply their 04:56:45.409 --> 04:56:50.340 exponents. That's exactly how we described the power rule above. It doesn't really matter 04:56:50.340 --> 04:56:56.819 if you multiply this exponent on the left side, or on the right side. But it's more 04:56:56.819 --> 04:57:03.189 traditional to multiply it on the left side. I've given these rules with the base of two, 04:57:03.189 --> 04:57:07.887 but they actually work for any base. To help you remember them, please take a moment to 04:57:07.887 --> 04:57:17.509 write out the log roles using a base of a you should get the following chart. Let's 04:57:17.509 --> 04:57:23.137 use the log rules to rewrite the following expressions as a sum or difference of logs. 04:57:23.137 --> 04:57:30.369 In the first expression, we have a log base 10 of a quotient. So we can rewrite the log 04:57:30.369 --> 04:57:39.307 of the quotient as the difference of the logs. Now we still have the log of a product, I 04:57:39.308 --> 04:57:47.727 can rewrite the log of a product as the sum of the logs. So that is log of y plus log 04:57:47.727 --> 04:57:55.099 of z. When I put things together, I have to be careful because here I'm subtracting the 04:57:55.099 --> 04:58:02.529 entire log expression. So I need to subtract both terms of this song on the make sure I 04:58:02.529 --> 04:58:08.270 do that by putting them in parentheses. Now I can simplify a little bit by distributing 04:58:08.270 --> 04:58:16.340 the negative sign. And here's my final answer. In my next expression, I have the log of a 04:58:16.340 --> 04:58:21.909 product. So I can rewrite that as the sum of two logs. 04:58:21.909 --> 04:58:31.042 I can also use my power rule to bring down the exponent T and multiply it in the front. 04:58:31.042 --> 04:58:38.909 That gives me the final expression log of five plus t times log of to one common mistake 04:58:38.909 --> 04:58:47.968 on this problem is to rewrite this expression as t times log of five times two. In fact, 04:58:47.968 --> 04:58:54.218 those two expressions are not equal. Because the T only applies to the two, not to the 04:58:54.218 --> 04:58:59.729 whole five times two, we can't just bring it down in front using the power wall. After 04:58:59.729 --> 04:59:08.349 all, the power rule only applies to a single expression raised to an exponent, and not 04:59:08.349 --> 04:59:14.457 to a product like this. And these next examples, we're going to go the other direction. Here 04:59:14.457 --> 04:59:18.729 we're given sums and differences of logs. And we want to wrap them up into a single 04:59:18.729 --> 04:59:24.989 log expression. By look at the first two pieces, that's a difference of logs. So I know I can 04:59:24.990 --> 04:59:34.978 rewrite it as the log of a quotient. Now I have the sum of two logs. So I can rewrite 04:59:34.977 --> 04:59:43.889 that as the log of a product. I'll clean that up a little bit and rewrite it as log base 04:59:43.889 --> 04:59:54.610 five of a times c over B. In my second example, I can rewrite the sum of my logs as the log 04:59:54.610 --> 05:00:03.369 of a product now, I will Like to rewrite this difference of logs as the log of a quotient. 05:00:03.369 --> 05:00:09.817 But I can't do it yet, because of that factor of two multiplied in front. But I can use 05:00:09.817 --> 05:00:15.669 the power rule backwards to put that two back up in the exponent. So I'll do that first. 05:00:15.669 --> 05:00:23.909 So I will copy down the ln of x plus one times x minus one, and rewrite this second term 05:00:23.909 --> 05:00:31.968 as ln of x squared minus one squared. Now I have a straightforward difference of two 05:00:31.968 --> 05:00:40.990 logs, which I can rewrite as the log of a quotient. I can actually simplify this some 05:00:40.990 --> 05:00:50.030 more. Since x plus one times x minus one is the same thing as x squared minus one. I can 05:00:50.029 --> 05:01:00.119 cancel factors to get ln of one over x squared minus one. In this video, we saw four rules 05:01:00.119 --> 05:01:07.329 for logs that are related to exponent rules. First, we saw that the log with any base of 05:01:07.330 --> 05:01:15.330 one is equal to zero. Second, we saw the product rule, the log of a product is equal to the 05:01:15.330 --> 05:01:23.968 sum of the logs. We saw the quotient rule, the log of a quotient is the difference of 05:01:23.968 --> 05:01:30.639 the logs. And we saw the power rule. When you take a log of an expression with an exponent 05:01:30.639 --> 05:01:37.200 in it, you can bring down the exponent and multiply it. It's worth noticing that there's 05:01:37.200 --> 05:01:44.137 no log rule that helps you split up the log of a psalm. In particular, the log of a psalm 05:01:44.137 --> 05:01:52.489 is not equal to the sum of the logs. If you think about logs and exponent rules going 05:01:52.490 --> 05:02:00.740 together, this kind of makes sense, because there's also no rule for rewriting the sum 05:02:00.740 --> 05:02:04.450 of two exponential expressions. 05:02:04.450 --> 05:02:10.308 Log rules will be super handy, as we start to solve equations using logs. Anytime you 05:02:10.308 --> 05:02:17.760 have an equation like this one that has variables in the exponent logs are the tool of choice 05:02:17.759 --> 05:02:23.387 for getting those variables down where you can solve for them. In this video, I'll do 05:02:23.387 --> 05:02:29.797 a few examples of solving equations with variables in the exponent. For our first example, let's 05:02:29.797 --> 05:02:37.250 solve for x the equation five times two to the x plus one equals 17. As my first step, 05:02:37.250 --> 05:02:42.707 I'm going to isolate the difficult spot, the part that has the variables and the exponent. 05:02:42.707 --> 05:02:47.889 In this example, I can do that by dividing both sides by five. That gives me two to the 05:02:47.889 --> 05:02:56.250 x plus one equals 17 over five. Next, I'm going to take the log of both sides, it's 05:02:56.250 --> 05:03:00.919 possible to take the log with any base, but I prefer to take the log base 10, or the log 05:03:00.919 --> 05:03:06.149 base e for the simple reason that my calculator has buttons for those logs. So in this example, 05:03:06.150 --> 05:03:12.610 I'll just take the log base 10. So I can omit the base because it's a base 10 is implied 05:03:12.610 --> 05:03:19.500 here. And that gives me this expression. As my next step, I'm going to use log roles to 05:03:19.500 --> 05:03:24.547 bring down my exponent and multiply it on the front. It's important to use parentheses 05:03:24.547 --> 05:03:31.217 here because the entire x plus one needs to get multiplied by log two. So that was my 05:03:31.218 --> 05:03:37.409 third step using the log roles. Now all my variables are down from the exponent where 05:03:37.409 --> 05:03:42.040 I can work with them, but I still need to solve for x. Right now x is trapped in the 05:03:42.040 --> 05:03:47.100 parentheses. So I'm going to free it from the parentheses by distributing. So I get 05:03:47.099 --> 05:03:55.887 x log two plus log two equals log of 17 fifths. Now I will try to isolate x by moving all 05:03:55.887 --> 05:04:01.520 my terms with x's in them to one side, and all my terms without x's in them to the other 05:04:01.520 --> 05:04:08.850 side. Finally, I factor out my x. Well, it's already kind of factored out, and I divide 05:04:08.849 --> 05:04:17.797 to isolate it. So I read out what I did. So far I distributed. I moved all the x terms 05:04:17.797 --> 05:04:24.739 on one side, and the terms without x's on the other side. And then I isolated the x 05:04:24.740 --> 05:04:31.830 by factoring out and dividing. We have an exact solution for x. This is correct, but 05:04:31.830 --> 05:04:36.830 maybe not so useful if you want a decimal answer. At this point, you can type in everything 05:04:36.830 --> 05:04:46.870 into your calculator using parentheses liberally to get a decimal answer about 0.765. It's 05:04:46.869 --> 05:04:51.829 always a good idea to check your work by typing in this decimal answer and seeing if the equation 05:04:51.830 --> 05:04:57.540 checks out. This next example is trickier because there are variables in the exponent 05:04:57.540 --> 05:05:05.180 in two places with two different bases. First step is normally to clean things up and simplify 05:05:05.180 --> 05:05:10.207 by isolating the tricky stuff. But in this example, there's nothing really to clean up 05:05:10.207 --> 05:05:15.590 or simplifier, or no way to isolate anything more than it already is. So we'll go ahead 05:05:15.590 --> 05:05:20.400 to the next step and take the log of both sides. Again, I'll go ahead and use log base 05:05:20.400 --> 05:05:27.548 10. But we couldn't use log base e instead. Next, we can use log roles to bring down the 05:05:27.547 --> 05:05:35.789 exponents. This gives me 2x minus three in parentheses, log two equals x minus two, log 05:05:35.790 --> 05:05:43.138 five. Now I'm going to distribute things out to free the excess from the parentheses. That 05:05:43.137 --> 05:05:52.950 gives me 2x log two minus three log two equals x log five minus two log five. Now I need 05:05:52.950 --> 05:05:59.350 to group the x terms on one side, and the other terms that don't involve x on the other 05:05:59.349 --> 05:06:08.099 side. So I'll keep the 2x log two on the left, put the minus log x log five on the left, 05:06:08.099 --> 05:06:13.759 and that gives me the minus two log five that stays on the right and a plus three log two 05:06:13.759 --> 05:06:22.799 on the right. Finally, I need to isolate x by factoring and dividing by factoring I just 05:06:22.799 --> 05:06:27.009 mean I factor out the x from all the terms on the left, so that gives me x times the 05:06:27.009 --> 05:06:37.989 quantity to log two minus log five. And that equals this mess on the right. Now I can divide 05:06:37.990 --> 05:06:48.540 the right side by the quantity on the left side. When you type this into your calculator, 05:06:48.540 --> 05:06:53.850 I encourage you to type the whole thing in rather than type bits and pieces in because 05:06:53.849 --> 05:06:58.159 if you round off, you'll get a less accurate answer than if you type the whole thing in 05:06:58.159 --> 05:07:06.648 at once. In this example, when I type it in, I get a final answer of about 5.106. 05:07:06.648 --> 05:07:12.670 In this equation, we have the variable t in the exponent in two places. The letter E is 05:07:12.669 --> 05:07:19.500 not a variable, it represents the number e whose decimal approximation is about 2.7. 05:07:19.500 --> 05:07:24.148 Because there's already an E and the expression, it's going to be handy to use natural log 05:07:24.148 --> 05:07:29.468 in this problem instead of log base 10. But before we take the log of both sides, let's 05:07:29.468 --> 05:07:35.468 clean things up. by isolating the tricky parts, we could at least divide both sides by five. 05:07:35.468 --> 05:07:42.409 And that will give us either the negative 0.05 t is equal to three fifths it is 0.2 05:07:42.409 --> 05:07:47.799 T. One way to proceed would now be to clean things out further by dividing both sides 05:07:47.799 --> 05:07:51.898 by either the point two t, but I'm going to take a different approach and go ahead and 05:07:51.898 --> 05:08:01.290 take the natural log of both sides. That gives me ln of e to the minus 0.05 t is equal to 05:08:01.290 --> 05:08:08.479 ln of three fifths e to the 0.2 t. Now on the left side, I can immediately use my log 05:08:08.479 --> 05:08:18.099 rule to bring down my exponent and get minus 0.05 t, L and E. But on the right side, I 05:08:18.099 --> 05:08:23.519 can't bring down the exponent yet because this e to the point two t is multiplied by 05:08:23.520 --> 05:08:28.797 three fifths. So before I can bring down the exponent, I need to split up this product 05:08:28.797 --> 05:08:36.110 using the product rule. So I can rewrite this as ln of three fifths plus ln e to the 0.2 05:08:36.110 --> 05:08:47.887 T. And now I can bring down the exponent. So that third step was using log roles to 05:08:47.887 --> 05:08:56.128 ultimately bring down my exponents. Now ln of E is a really nice expression, ln IV means 05:08:56.128 --> 05:09:02.431 log base e of E. So that's asking what power do I raise E to in order to get e? And the 05:09:02.431 --> 05:09:10.570 answer is one. So anytime I have ln of E, I can just replace that with one. That's why 05:09:10.570 --> 05:09:14.990 using natural log is a little bit handier here than using log base 10. You can make 05:09:14.990 --> 05:09:23.770 that simplification. Next, I'm ready to solve for t. So I don't need to distribute here. 05:09:23.770 --> 05:09:29.610 But I do need to bring my T terms to one side of my terms without t to the other side. So 05:09:29.610 --> 05:09:37.779 let's see. I'll put my T terms on the left and my team turns without T on the right. 05:09:37.779 --> 05:09:45.619 And finally I'm going to isolate t by factoring and dividing. So by factoring I mean I factor 05:09:45.619 --> 05:09:57.147 out my T and now I can divide. Using my calculator, 05:09:57.148 --> 05:10:04.718 I can get a decimal answer of 2.04 Three, three. This video gave some examples of solving 05:10:04.718 --> 05:10:10.477 equations with variables in the exponent. And the key idea was to take the log of both 05:10:10.477 --> 05:10:21.270 sides and use the log properties to bring the exponents down. This video, give some 05:10:21.270 --> 05:10:26.718 examples of equations with logs in them like this one. In order to solve the equations 05:10:26.718 --> 05:10:32.540 like this one, we have to free the variable from the log. And we'll do that using exponential 05:10:32.540 --> 05:10:38.830 functions. My first step in solving pretty much any kind of equation is to simplify it 05:10:38.830 --> 05:10:44.000 and isolate the tricky part. In this case, the tricky part is the part with the log in 05:10:44.000 --> 05:10:51.599 it. So I can isolate it by first adding three to both sides. That gives me two ln 2x plus 05:10:51.599 --> 05:11:01.180 five equals four, and then I can divide both sides by two. Now that I've isolated the tricky 05:11:01.180 --> 05:11:07.520 part, I still need to solve for x, but x is trapped inside the log function. to free it, 05:11:07.520 --> 05:11:13.200 I need to somehow undo the log function. Well log functions and exponential functions undo 05:11:13.200 --> 05:11:18.727 each other. And since this is a log base e, I need to use the exponential function base 05:11:18.727 --> 05:11:26.270 e also. So I'm going to take e to the power of both sides. In other words, I'll take e 05:11:26.270 --> 05:11:33.420 to the ln 2x plus five, and that's going to equal e to the power of two. 05:11:33.419 --> 05:11:40.189 Now e to the ln of anything, I'll just write a for anything. That means e to the log base 05:11:40.189 --> 05:11:46.859 e of a, that you the power and log base e undo each other. So we just get a, I'm going 05:11:46.860 --> 05:11:53.200 to use that principle over here, e to the ln of 2x plus five, the E and the log base 05:11:53.200 --> 05:11:59.030 e undo each other. And we're left with 2x plus five on the left side. So 2x plus five 05:11:59.029 --> 05:12:04.360 is equal to E squared. And from there, it's easy to finish the problem and solve for x 05:12:04.360 --> 05:12:11.047 by subtracting five from both sides and then dividing by two. So I'll just write the third 05:12:11.047 --> 05:12:17.967 step here is to just say finish solving for x. There's one last step we need to do when 05:12:17.968 --> 05:12:25.909 solving equations with logs in them. And that's to check our answers. Because we may get extraneous 05:12:25.909 --> 05:12:31.529 solutions. an extraneous solution is a solution that comes out of the solving process, but 05:12:31.529 --> 05:12:36.807 doesn't actually satisfy the original equation. And those can happen for equations with logs 05:12:36.808 --> 05:12:42.317 in them, because we might get a solution that makes the argument of the log negative or 05:12:42.317 --> 05:12:52.898 zero, and we can't take the log of a negative numbers zero. So let's check and plug in the 05:12:52.898 --> 05:13:01.020 solution of E squared minus five over two. We'll plug that in for x in our original equation 05:13:01.020 --> 05:13:09.840 and see if that works. So let's see the twos cancel here. So I get two ln e squared minus 05:13:09.840 --> 05:13:17.049 five plus five, minus three, I want that to equal one. And now my fives cancel. And so 05:13:17.049 --> 05:13:23.648 I have two ln e squared minus three that I want to equal one. Well, I think this is going 05:13:23.649 --> 05:13:34.900 to work out because let's see, ln is log base e. and log base e of E squared that's asking 05:13:34.900 --> 05:13:38.890 the question, What power do I raise e two to get e squared? Well, I have to raise it 05:13:38.889 --> 05:13:46.750 to the power of two to get e squared. So this becomes two times two minus three. And does 05:13:46.750 --> 05:13:53.080 that equal one, four minus three does equal one. So that all checks out. And we didn't 05:13:53.080 --> 05:13:58.780 have any problem with taking the log of negative numbers zero, we didn't get any extraneous 05:13:58.779 --> 05:14:04.619 solutions. So this is our solution. The second equation is a little bit trickier, because 05:14:04.619 --> 05:14:10.637 there's a log into places. Now notice that this is a log where there's no base written. 05:14:10.637 --> 05:14:16.340 So a base 10 is implied. So I'm already thinking in order to undo a log base 10, I'm going 05:14:16.340 --> 05:14:23.080 to want to take a 10 to the power of both sides. Of course, it's still a good idea to 05:14:23.080 --> 05:14:29.060 isolate the tricky part, but there's nothing really to isolate here. So I'm going to say, 05:14:29.060 --> 05:14:37.298 can't do it here. So we'll just jump right to straight step two, and take 10 to the power 05:14:37.297 --> 05:14:46.887 of both sides. Okay, so that's going to give me 10 to the whole thing, log x plus three 05:14:46.887 --> 05:14:52.389 plus log x, that whole thing is in the exponent equals 10 to the One Power. Well, I know what 05:14:52.389 --> 05:14:58.750 to do with the right side 10 to the one is just 10. But what do I do with the 10 to these 05:14:58.750 --> 05:14:59.840 two things added 05:14:59.840 --> 05:15:00.930 up Up, 05:15:00.930 --> 05:15:06.869 while remembering my exponent rules, I know that when you add up the exponent, that's 05:15:06.869 --> 05:15:12.419 what happens when you multiply two things. So this is the same thing as 10 to the log 05:15:12.419 --> 05:15:18.369 x plus three times 10 to the log x, right, because when you multiply two things, you 05:15:18.369 --> 05:15:24.439 add the exponent, so these are the same. Okay, now we're in business, because 10 to the log 05:15:24.439 --> 05:15:31.797 base 10, those undo each other. And so this whole expression here simplifies to x plus 05:15:31.797 --> 05:15:38.967 three. Similarly, 10 to the log base, 10 of x is just x. So I'm multiplying x plus three 05:15:38.968 --> 05:15:46.030 by x, that's equal to 10. Now I have an equation I can deal with, it's a quadratic, so I'm 05:15:46.029 --> 05:15:52.397 going to first multiply out to make it look more like a quadratic, get everything to one 05:15:52.398 --> 05:15:59.409 side. So is equal to zero. And, and now I can either factor or use the quadratic formula. 05:15:59.409 --> 05:16:06.500 I think this one factors, it looks like X plus five times x minus two, so I'm going 05:16:06.500 --> 05:16:14.060 to get x is negative five, or x is two. So that was all the third step to finish solving 05:16:14.060 --> 05:16:20.860 for x. Finally, we need to check our solutions to make sure we haven't gotten some extraneous 05:16:20.860 --> 05:16:27.378 ones. So let's see if x equals negative five. If I plug that into my original equation, 05:16:27.378 --> 05:16:36.047 that says, I'm checking that log of negative five plus three, plus log of negative five, 05:16:36.047 --> 05:16:40.879 checking that's equal to one. Well, this is giving me a queasy feeling. And I hope it's 05:16:40.880 --> 05:16:48.540 giving you a queasy feeling too, because log of negative two does not exist, right, you 05:16:48.540 --> 05:16:53.750 can't take the log of a negative number. Same thing with log of negative five. So x equals 05:16:53.750 --> 05:17:00.297 negative five is an extraneous solution, it doesn't actually solve our original equation. 05:17:00.297 --> 05:17:08.029 Let's check the other solution, x equals two. So now we're checking to see if log of two 05:17:08.029 --> 05:17:15.360 plus three plus log of two is equal to one. Since there's no problem with taking logs 05:17:15.360 --> 05:17:20.670 of negative numbers, or zero here, this should work out fine. And just to finish checking, 05:17:20.669 --> 05:17:27.423 we can see let's see this is log of five plus log of two, we want that equal one. But using 05:17:27.423 --> 05:17:34.540 my log rules, the sum of two logs is the log of the product. So log of five times two, 05:17:34.540 --> 05:17:41.148 we want that to equal one. And that's just log base 10 of 10. And that definitely equals 05:17:41.148 --> 05:17:46.909 one because log base 10 of 10 says, What power do I raise tend to to get 10. And that power 05:17:46.909 --> 05:17:56.079 is one. So the second solution x equals two does check out. And that's our final answer. 05:17:56.080 --> 05:18:00.350 Before I leave this problem, I do want to mention that some people have an alternative 05:18:00.349 --> 05:18:05.659 approach. Some people like to start with the original equation, and then use log roles 05:18:05.659 --> 05:18:12.790 to combine everything into one log expression. So since we have the sum of two logs, we know 05:18:12.790 --> 05:18:18.388 that's the same as the log of a product, right, so we can rewrite the left side as log of 05:18:18.387 --> 05:18:25.849 x plus three times x, that equals one, then we do the same trick of taking 10 to the power 05:18:25.849 --> 05:18:32.717 of both sides. And as before, the 10 to the power and the log base 10 undo each other, 05:18:32.718 --> 05:18:40.690 and we get x plus three times x equals 10, just like we did before. In the first solution, 05:18:40.689 --> 05:18:45.639 we ended up using exponent rules to rewrite things. In the second alternative method, 05:18:45.639 --> 05:18:50.689 we use log rules to rewrite things. So the methods are really pretty similar, pretty 05:18:50.689 --> 05:18:54.849 equivalent, and they certainly will get us to the same answer. So we've seen a couple 05:18:54.849 --> 05:19:02.137 examples of equations with logs in them and how to solve them. And the key step is to 05:19:02.137 --> 05:19:13.279 use exponential functions to undo the log. In other words, take e to the power of both 05:19:13.279 --> 05:19:25.090 sides to undo natural log, and take 10 to the power of both sides. To undo log base 05:19:25.090 --> 05:19:27.560 10. 05:19:27.560 --> 05:19:33.030 In this video, we'll use exponential equations in some real life applications, like population 05:19:33.029 --> 05:19:40.217 growth and radioactive decay. I'll also introduce the ideas of half life and doubling time. 05:19:40.218 --> 05:19:47.350 In this first example, let's suppose we invest $1,600 in a bank account that earns 6.5% annual 05:19:47.349 --> 05:19:52.889 interest compounded once a year. How many years will it take until the account has $2,000 05:19:52.889 --> 05:19:59.450 in it if you don't make any further deposits or withdrawals since our money is earning 05:19:59.450 --> 05:20:05.540 six point 5% interest each year, that means that every year the money gets multiplied 05:20:05.540 --> 05:20:18.817 by 1.065. So after t years, my 1600 gets multiplied by 1.065 to the t power. I'll write this in 05:20:18.817 --> 05:20:30.308 function notation as f of t equals 1600 times 1.065 to the T, where f of t is the amount 05:20:30.308 --> 05:20:41.080 of money after t years. Now we're trying to figure out how long it will take to get $2,000 05:20:41.080 --> 05:20:48.580 $2,000 is a amount of money. So that's an amount for f of t. And we're trying to solve 05:20:48.580 --> 05:20:55.290 for T the amount of time. So let me write out my equation and make a note that I'm solving 05:20:55.290 --> 05:21:04.840 for t. Now to solve for t, I want to first isolate the tricky part. So I'm going to the 05:21:04.840 --> 05:21:08.790 tricky part is the part with the exponential in it. So I'm going to divide both sides by 05:21:08.790 --> 05:21:18.190 1600. That gives me 2000 over 1600 equals 1.065. To the tee, I can simplify this a little 05:21:18.189 --> 05:21:24.317 bit further as five fourths. Now that I've isolated the tricky part, my next step is 05:21:24.317 --> 05:21:29.727 going to be to take the log of both sides. That's because I have a variable in the exponent. 05:21:29.727 --> 05:21:33.829 And I know that if I log take the log of both sides, I can use log roles to bring that exponent 05:21:33.830 --> 05:21:40.850 down where I can solve for it. I think I'll use log base e this time. So I have ln fi 05:21:40.849 --> 05:21:48.759 force equals ln of 1.065 to the T. Now by the power rule for logs, on the right side, 05:21:48.759 --> 05:21:56.919 I can bring that exponent t down and multiply it in the front. Now it's easy to isolate 05:21:56.919 --> 05:22:07.329 t just by dividing both sides by ln of 1.065. Typing that into my calculator, I get that 05:22:07.330 --> 05:22:17.308 t is approximately 3.54 years. And the next example, we have a population of bacteria 05:22:17.308 --> 05:22:24.290 that initially contains 1.5 million bacteria, and it's growing by 12% per day, we want to 05:22:24.290 --> 05:22:34.659 find the doubling time, the doubling time means the amount of time it takes for a quantity 05:22:34.659 --> 05:22:41.110 to double in size. For example, the amount of time it takes to get from the initial 1.5 05:22:41.110 --> 05:22:47.540 million bacteria to 3 million bacteria would be the doubling time. Let's start by writing 05:22:47.540 --> 05:22:56.558 an equation for the amount of bacteria. So if say, P of t represents the number of bacteria 05:22:56.558 --> 05:23:08.968 in millions, then my equation and T represents time in days, then P of t is going to be given 05:23:08.968 --> 05:23:19.887 by the initial amount of bacteria times the growth factor 1.12 to the T. That's because 05:23:19.887 --> 05:23:25.360 my population of bacteria is growing by 12% per day. That means the number of bacteria 05:23:25.360 --> 05:23:30.760 gets multiplied by one point 12. Since we're looking for the doubling time, we're looking 05:23:30.759 --> 05:23:39.309 for the t value when P of D will be twice as big. So I can set P of t to be three and 05:23:39.310 --> 05:23:56.298 solve for t. As before, I'll start by isolating the tricky part, taking the log 05:23:56.297 --> 05:24:08.237 bringing the T down. And finally solving for T. Let's see three over 1.5 is two so I can 05:24:08.238 --> 05:24:19.450 write this as ln two over ln 1.12. Using my calculator, that's about 6.12 days. It's an 05:24:19.450 --> 05:24:32.387 interesting fact that doubling time only depends on the growth rate, that 12% growth, not the 05:24:32.387 --> 05:24:37.137 initial population. In fact, I could have figured out the doubling time without even 05:24:37.137 --> 05:24:41.647 knowing how many bacteria were in my initial population. Let me show you how that would 05:24:41.648 --> 05:24:48.100 work. If I didn't know how many I started with, I could still write P of t equals a 05:24:48.099 --> 05:24:55.989 times 1.12 to the t where a is our initial population that I don't know what it is. 05:24:55.990 --> 05:25:01.121 Then if I want to figure out how long it takes for my population, to double Well, if I start 05:25:01.120 --> 05:25:08.869 with a and double that I get to a. So I'll set my population equal to two a. And I'll 05:25:08.869 --> 05:25:19.297 solve for t. Notice that my A's cancel. And so when I take the log of both sides and bring 05:25:19.297 --> 05:25:32.099 the T down and solve for t, I get the exact same thing as before, it didn't matter what 05:25:32.099 --> 05:25:38.009 the initial population was, I didn't even have to know what it was. In this next example, 05:25:38.009 --> 05:25:42.217 we're told the initial population, and we're told the doubling time, we're not told by 05:25:42.218 --> 05:25:49.218 what percent the population increases each minute, or by what number, we're forced to 05:25:49.218 --> 05:25:53.440 multiply the population by each minute. So we're gonna have to solve for that, I do know 05:25:53.439 --> 05:25:59.340 that I want to use an equation of the form y equals a times b to the T, where T is going 05:25:59.340 --> 05:26:04.920 to be the number of minutes, and y is going to be the number of bacteria. And I know that 05:26:04.919 --> 05:26:11.459 my initial amount a is 350. So I can really write y equals 350 times b to the T. Now the 05:26:11.459 --> 05:26:19.119 doubling time tells me that when 15 minutes have elapsed, my population is going to be 05:26:19.119 --> 05:26:28.619 twice as big, or 700. plugging that into my equation, I have 700 equals 350 times b to 05:26:28.619 --> 05:26:34.789 the 15. Now I need to solve for b. Let me clean this up a little bit by dividing both 05:26:34.790 --> 05:26:43.798 sides by 350. That gives me 700 over 350 equals b to the 15th. In other words, two equals 05:26:43.797 --> 05:26:51.479 b to the 15th. To solve for B, I don't have to actually use logs here, because my variables 05:26:51.479 --> 05:26:57.270 in the base not in the exponent, so I don't need to bring that exponent down. Instead, 05:26:57.270 --> 05:27:03.439 the easiest way to solve this is just by taking the 15th root of both sides or equivalently. 05:27:03.439 --> 05:27:12.467 The 1/15 power. That's because if I take B to the 15th to the 1/15, I multiply by exponents, 05:27:12.468 --> 05:27:19.690 that gives me B to the one is equal to two to the 1/15. In other words, B is two to the 05:27:19.689 --> 05:27:29.789 1/15, which as a decimal is approximately 1.047294. I like to use a lot of decimals 05:27:29.790 --> 05:27:34.170 if I'm doing a decimal approximation and these kind of problems to increase accuracy. But 05:27:34.169 --> 05:27:38.467 of course, the most accurate thing is just to leave B as it is. And I'll do that and 05:27:38.468 --> 05:27:50.190 rewrite my equation is y equals 350 times two to the 1/15 to the T. Now I'd like to 05:27:50.189 --> 05:27:56.079 work this problem one more time. And this time, I'm going to use the form of the equation 05:27:56.080 --> 05:28:04.370 y equals a times e to the RT. This is called a continuous growth model. It looks different, 05:28:04.369 --> 05:28:09.750 but it's actually an equivalent form to this growth model over here. And I'll say more 05:28:09.750 --> 05:28:14.450 about why these two forms are equivalent at the end, I can use the same general ideas 05:28:14.450 --> 05:28:21.950 to solve in this form. So I know that my initial amount is 350. And I know again that when 05:28:21.950 --> 05:28:33.530 t is 15, my Y is 700. So I plug in 700 here 350 e to the r times 15. And I solve for r. 05:28:33.529 --> 05:28:44.869 Again, I'm going to simplify things by dividing both sides by 350. That gives me two equals 05:28:44.869 --> 05:28:51.840 e to the r times 15. This time my variable does end up being in the exponent. So I do 05:28:51.840 --> 05:28:56.830 want to take the log of both sides, I'm going to use natural log, because I already have 05:28:56.830 --> 05:29:01.818 an E and my problem. So natural log and E are kind of more harmonious the other than 05:29:01.817 --> 05:29:09.259 then common log with base 10 and E. So I take the log of both sides. Now I can pull the 05:29:09.259 --> 05:29:14.789 exponent down. So that's our times 15 times ln of E. Well Elena V is just one, right? 05:29:14.790 --> 05:29:19.540 Because Elena V is asking what power do I raise E to to get e that's just one. And so 05:29:19.540 --> 05:29:27.218 I get 15 r equals ln two. So r is equal to ln two divided by 15. Let me plug that back 05:29:27.218 --> 05:29:35.250 in to my original equation as e to the ln two over 15 times t. Now I claimed that these 05:29:35.250 --> 05:29:40.279 two equations were actually the same thing just looking different. And the way to see 05:29:40.279 --> 05:29:49.099 that is if I start with this equation, and I rewrite as e to the ln two over 15 like 05:29:49.099 --> 05:29:50.099 that 05:29:50.099 --> 05:29:58.090 to the tee. Well, I claim that this quantity right here is the same thing as my two to 05:29:58.090 --> 05:30:06.727 the 1/15. And in fact One way to see that is e to the ln two already to the 115. The 05:30:06.727 --> 05:30:10.509 Right, that's the same, because every time I take a power to a power, I multiply exponents. 05:30:10.509 --> 05:30:18.840 But what's Ed Oh, and two, E and ln undo each other, so that's just 350 times two to the 05:30:18.840 --> 05:30:24.887 1/15 to the T TA, the equations are really the same. And these, you'll always be able 05:30:24.887 --> 05:30:30.897 to find two different versions of an exponential equation, sort of the standard one, a times 05:30:30.898 --> 05:30:38.738 b to the T, or the continuous growth one, a times e to the RT. In this last example, 05:30:38.738 --> 05:30:43.569 we're going to work with half life, half life is pretty much like doubling time, it just 05:30:43.569 --> 05:30:55.099 means the amount of time that it takes for a quantity to decrease to half as much as 05:30:55.099 --> 05:31:02.509 we originally started with, we're told that the half life of radioactive carbon 14 is 05:31:02.509 --> 05:31:11.009 5750 years. So that means it takes that long for a quantity of radioactive carbon 14 to 05:31:11.009 --> 05:31:16.807 decay, so that you just have half as much left and the rest is nonradioactive form. 05:31:16.808 --> 05:31:21.718 So we're told a sample of bone that originally contained 200 grams of radioactive carbon 05:31:21.718 --> 05:31:29.670 14 now contains only 40 grams, we're supposed to find out how old the sample is, is called 05:31:29.669 --> 05:31:36.957 carbon dating. Let's use the continuous growth model this time. So our final amount, so this 05:31:36.957 --> 05:31:49.099 is our amount of radioactive c 14 is going to be the initial amount 05:31:49.099 --> 05:31:54.250 times e to the RT, we could have used the other model to we could have used f of t equals 05:31:54.250 --> 05:31:59.468 a times b to the T, but I just want to use the continuous model for practice. So we know 05:31:59.468 --> 05:32:13.659 that our half life is 5750. So what that means is when t is 5750, our amount is going to 05:32:13.659 --> 05:32:19.290 be one half of what we started with. Let me see if I can plug that into my equation and 05:32:19.290 --> 05:32:26.968 figure out use that to figure out what r is. That's ours called the continuous growth rate. 05:32:26.968 --> 05:32:35.350 So I plug in one half a for the final amount, a is still the initial amount, e to the r 05:32:35.349 --> 05:32:46.279 and I have 5750 I can cancel my A's. And now I want to solve for r, r is in my exponent. 05:32:46.279 --> 05:32:50.987 So I do need to take the log of both sides to solve for it. I'm going to use log base 05:32:50.988 --> 05:32:57.120 e since I already have an E and my problem, log base e is more compatible with E then 05:32:57.119 --> 05:33:04.227 log base 10 is okay. Now, on the left side, I still have log of one half and natural log 05:33:04.227 --> 05:33:09.887 of one half on the right side, ln n e to a power those undo each other. So I'm left with 05:33:09.887 --> 05:33:20.649 R times 5750. Now I can solve for r, it's ln one half over 5750 could work that as a 05:33:20.650 --> 05:33:25.909 decimal, but it's actually more accurate just to keep it in exact form. So now I can rewrite 05:33:25.909 --> 05:33:40.707 my equation, I have f of t equals a times e to the ln one half over five 750 t. Now 05:33:40.707 --> 05:33:46.750 I can use that to figure out my problem. And my problem the bone originally contained 200 05:33:46.750 --> 05:33:53.468 grams, that's my a, I want to figure out when it's going to contain only 40 grams, that's 05:33:53.468 --> 05:34:00.030 my final amount. And so I need to solve for t. I'll clean things up by dividing both sides 05:34:00.029 --> 05:34:10.501 by 200. Let's say 40 over 200 is 1/5. Now I'm going to take the ln of both sides. And 05:34:10.501 --> 05:34:19.029 ln and e to the power undo each other. So I'm left with Elena 1/5 equals ln of one half 05:34:19.029 --> 05:34:27.889 divided by five 750 T. And finally I can solve for t but super messy but careful use of my 05:34:27.889 --> 05:34:35.509 calculator gives me an answer of 13,351 years approximately. 05:34:35.509 --> 05:34:41.397 That kind of makes sense in terms of the half life because to get from 200 to 40 you have 05:34:41.398 --> 05:34:45.638 to decrease by half a little more than two times right or increasing by half once would 05:34:45.637 --> 05:34:50.659 get you to 100 decreasing to half again would get you to 50 a little more than 40 and two 05:34:50.659 --> 05:34:57.349 half lives is getting pretty close to 13,000 years. This video introduced a lot of new 05:34:57.349 --> 05:35:04.797 things it introduced continue Less growth model, which is another equivalent way of 05:35:04.797 --> 05:35:16.939 writing an exponential function. The relationship is that the B in this example, is the same 05:35:16.939 --> 05:35:27.619 thing as EDR. In that version, it also introduced the ideas of doubling time. And HalfLife the 05:35:27.619 --> 05:35:32.409 amount of time it takes a quantity to double or decreased to half in an exponential growth 05:35:32.409 --> 05:35:42.701 model. Recall that a linear equation is equation like for example, 2x minus y equals one. It's 05:35:42.701 --> 05:35:47.990 an equation without any x squared or y squared in it, something that could be rewritten in 05:35:47.990 --> 05:35:55.120 the form y equals mx plus b, the equation for a line a system of linear equations is 05:35:55.119 --> 05:36:03.669 a collection of two or more linear equations. For example, I could have these two equations. 05:36:03.669 --> 05:36:10.089 A solution to a system of equations is that an x value and a y value that satisfy both 05:36:10.090 --> 05:36:18.590 of the equations. For example, the ordered pair two three, that means x equals two, y 05:36:18.590 --> 05:36:26.950 equals three is a solution to this system. Because if I plug in x equals two, and y equals 05:36:26.950 --> 05:36:33.968 three into the first equation, it checks out since two times two minus three is equal to 05:36:33.968 --> 05:36:40.170 one. And if I plug in x equals two and y equals three into the second equation, it also checks 05:36:40.169 --> 05:36:49.279 out two plus three equals five. However, the ordered pair one for that is x equals one, 05:36:49.279 --> 05:36:55.849 y equals four is not a solution to the system. Even though this X and Y value work in the 05:36:55.849 --> 05:37:02.707 second equation, since one plus four does equal five, it doesn't work in the first equation, 05:37:02.707 --> 05:37:10.567 because two times one minus four is not equal to one. In this video, we'll use systematic 05:37:10.567 --> 05:37:17.229 methods to find the solutions to systems of linear equations. In this first example, we 05:37:17.229 --> 05:37:22.181 want to solve this system of equations, there are two main methods we could use, we could 05:37:22.181 --> 05:37:29.900 use the method of substitution, or we could use the method of elimination. If we use the 05:37:29.900 --> 05:37:38.128 method of substitution, the main idea is to isolate one variable in one equation, and 05:37:38.128 --> 05:37:44.440 then substitute it in to the other equation. For example, we can start with the first equation 05:37:44.439 --> 05:37:53.807 3x minus two y equals four, and isolate the x by adding two y to both sides, and then 05:37:53.808 --> 05:38:01.080 dividing both sides by three. Think I'll rewrite that a little bit by breaking up the fraction 05:38:01.080 --> 05:38:07.900 into two fractions four thirds plus two thirds y. Now, I'm going to copy down the second 05:38:07.900 --> 05:38:16.530 equation 5x plus six y equals two. And I'm going to substitute in my expression for x. 05:38:16.529 --> 05:38:24.887 That gives me five times four thirds plus two thirds y plus six y equals two. And now 05:38:24.887 --> 05:38:30.750 I've got an equation with only one variable in it y. So I can solve for y as a number. 05:38:30.750 --> 05:38:38.830 First, I'm going to distribute the five so that gives me 20 thirds plus 10 thirds y plus 05:38:38.830 --> 05:38:45.548 six y equals two. And now I'm going to keep all my y terms, my terms with y's in them 05:38:45.547 --> 05:38:51.547 on the left side, but I'll move all my terms without y's in them to the right side. At 05:38:51.547 --> 05:38:56.110 this point, I could just add up all my fractions and solve for y. But since I don't really 05:38:56.110 --> 05:39:00.360 like working with fractions, I think I'll do the trick of clearing the denominators 05:39:00.360 --> 05:39:05.968 here. So I'm going to actually multiply both sides by my common denominator of three just 05:39:05.968 --> 05:39:10.637 to get rid of the denominators and not have to work with fractions. So let me write that 05:39:10.637 --> 05:39:20.237 down. Distributing the three, I get 10 y plus eight t and y equals six minus 20. Now add 05:39:20.238 --> 05:39:28.290 things together. So that's 28 y equals negative 14. So that means that y is going to be negative 05:39:28.290 --> 05:39:32.110 14 over 28, which is negative one half. 05:39:32.110 --> 05:39:38.680 So I've solved for y. And now I can go back and plug y into either of my equations to 05:39:38.680 --> 05:39:47.977 solve for x, I plug it into my first equation. So I'm plugging in negative one half for y. 05:39:47.977 --> 05:39:55.058 That gives me 3x plus one equals four. So 3x equals three, which means that x is equal 05:39:55.058 --> 05:40:01.370 to one. I've solved my system of equations and gotten x equals one On y equals minus 05:40:01.369 --> 05:40:09.680 one half, I can also write that as an ordered pair, one, negative one half for my solution. 05:40:09.680 --> 05:40:14.297 Now let's go back and solve the same system, but use a different method, the method of 05:40:14.297 --> 05:40:23.750 elimination, the key idea to the method of elimination is to multiply each equation by 05:40:23.750 --> 05:40:34.718 a constant to make the coefficients of one variable match. Let me start by copying down 05:40:34.718 --> 05:40:42.080 my two equations. Say I'm trying to make the coefficient of x match. One way to do that 05:40:42.080 --> 05:40:48.730 is to multiply the first equation by five, and the second equation by three. That way, 05:40:48.729 --> 05:40:55.739 the coefficient of x will be 15 for both equations, so let me do that. So for the first equation, 05:40:55.740 --> 05:41:00.950 I'm going to multiply both sides by five. And for the second equation, I'm going to 05:41:00.950 --> 05:41:09.979 multiply both sides by three. That gives me for the first equation 15x minus 10, y equals 05:41:09.979 --> 05:41:19.110 20. And for the second equation, 15x plus 18, y is equal to six. Notice that the equate 05:41:19.110 --> 05:41:25.378 the coefficients of x match. So if I subtract the second equation from the first, the x 05:41:25.378 --> 05:41:32.779 term will completely go away, it'll be zero times x, and I'll be left with let's see negative 05:41:32.779 --> 05:41:44.259 10 y minus 18, y is going to give me minus 28 y. And if I do 20 minus six, that's going 05:41:44.259 --> 05:41:52.237 to give me 14. solving for y, I get y is 14 over minus 28, which is minus one half just 05:41:52.238 --> 05:41:59.228 like before. Now we can continue, like we did in the previous solution, and substitute 05:41:59.227 --> 05:42:06.290 that value of y into either one of the equations. I'll put it again in here. And my solution 05:42:06.290 --> 05:42:13.280 proceeds as before. So once again, I get the solution that x equals one and y equals minus 05:42:13.279 --> 05:42:20.599 one half. Before we go on to the next problem, let me show you graphically what this means. 05:42:20.599 --> 05:42:29.147 Here I've graphed the equations 3x minus two, y equals four, and 5x plus six y equals two. 05:42:29.148 --> 05:42:36.488 And we can see that these two lines intersect in the point with coordinates one negative 05:42:36.488 --> 05:42:42.388 one half, just like we predicted by solving equations algebraically. Let's take a look 05:42:42.387 --> 05:42:49.297 at another system of equations. I'm going to rewrite the first equation, so the x term 05:42:49.297 --> 05:42:55.329 is on the left side with the y term, and the constant term stays on the right. And I'll 05:42:55.330 --> 05:43:02.580 rewrite or copy down the second equation. Since the coefficient of x in the first equation 05:43:02.580 --> 05:43:08.270 is minus four, and the second equation is three, I'm going to try using the method of 05:43:08.270 --> 05:43:14.119 elimination and multiply the first equation by three and the second equation by four. 05:43:14.119 --> 05:43:19.009 That'll give me a coefficient of x of negative 12. And the first equation, and 12. And the 05:43:19.009 --> 05:43:23.799 second equation, those are equal and opposite, right, so I'll be able to add together my 05:43:23.799 --> 05:43:31.619 equations to cancel out access. So let's do that. My first equation becomes negative 12x 05:43:31.619 --> 05:43:39.009 plus 24, y equals three, and I'll put everything by three. And my second equation, I'll multiply 05:43:39.009 --> 05:43:46.859 everything by four. So that's 12x minus 24, y equals eight. Now something kind of funny 05:43:46.860 --> 05:43:53.369 has happened here, not only do the x coefficients match has in with opposite signs, but the 05:43:53.369 --> 05:44:00.169 Y coefficients do also. So if I add together my two equations, in order to cancel out the 05:44:00.169 --> 05:44:06.457 x term, I'm also going to cancel out the y term, and I'll just get zero plus zero is 05:44:06.457 --> 05:44:12.369 equal to three plus eight is 11. Well, that's a contradiction, we can't have zero equal 05:44:12.369 --> 05:44:19.817 to 11. And that shows that these two equations actually have no solution. 05:44:19.817 --> 05:44:25.628 Let's look at this situation graphically. If we graph our two equations, we see that 05:44:25.628 --> 05:44:31.440 they're parallel lines with the same slope. This might be more clear, if I isolate y in 05:44:31.439 --> 05:44:36.727 each equation, the first equation, I get y equals, let's see, dividing by eight that's 05:44:36.727 --> 05:44:43.329 the same thing as four eighths or one half x plus one eight. And the second equation, 05:44:43.330 --> 05:44:50.760 if I isolate y, let's say minus six y equals minus 3x plus two divided by minus six, that's 05:44:50.759 --> 05:44:59.487 y equals one half x minus 1/3. So indeed, they have the same slope. And so they're parallel 05:44:59.488 --> 05:45:04.409 with different In intercepts, and so they can have no intersection. And so it makes 05:45:04.409 --> 05:45:10.729 sense that we have no solution to our system of linear equations. This kind of system that 05:45:10.729 --> 05:45:18.298 has no solution is called an inconsistent system. In this third example, yet a third 05:45:18.298 --> 05:45:24.797 behavior happens. This time, I think I'm going to solve by substitution because I already 05:45:24.797 --> 05:45:30.579 have X with a coefficient of one. So it's really easy to just isolate X in the first 05:45:30.580 --> 05:45:39.228 equation, and then plug in to the second equation to get three times six minus five y plus 15, 05:45:39.227 --> 05:45:46.647 y equals 18. If I distribute out, I get the strange phenomenon that the 15 y's cancel 05:45:46.648 --> 05:45:54.488 and I just get 18 equals 18, which is always true. This is what's called a dependent system 05:45:54.488 --> 05:46:00.850 of linear equations. If you look more closely, you can see that the second equation is really 05:46:00.849 --> 05:46:06.169 just a constant multiple, the first equation is just three times every term is three times 05:46:06.169 --> 05:46:10.939 as big as the corresponding term and the first equation. So there's no new information in 05:46:10.939 --> 05:46:16.539 the second equation, anything, any x and y values that satisfy the first one will satisfy 05:46:16.540 --> 05:46:23.250 the second one. So this system of equations has infinitely many solutions. Any ordered 05:46:23.250 --> 05:46:32.580 pair x y, where X plus five y equals six, or in other words, X's minus five y plus six 05:46:32.580 --> 05:46:40.318 will satisfy this system of equations. That would include a y value of zero corresponding 05:46:40.317 --> 05:46:49.477 x value of six or a y value of one corresponding to an x value of one, or a y value of 1/3. 05:46:49.477 --> 05:46:55.259 Corresponding to an x value of 13 thirds just by plugging into this equation will work. 05:46:55.259 --> 05:47:00.679 Graphically, if I graph both of these equations, the lines will just be on top of each other, 05:47:00.680 --> 05:47:06.099 so I'll just see one line. In this video, we've solved systems of linear equations, 05:47:06.099 --> 05:47:13.199 using the method of substitution and the method of elimination. We've seen that systems of 05:47:13.200 --> 05:47:19.317 linear equations can have one solution. When the lines that the equations represent intersect 05:47:19.317 --> 05:47:26.797 in one point, they can be inconsistent, and have no solutions that corresponds to parallel 05:47:26.797 --> 05:47:33.029 lines, or they can be dependent and have infinitely many solutions that corresponds to the lines 05:47:33.029 --> 05:47:39.647 lying on top of each other. In this video, I'll work through a problem involving distance 05:47:39.648 --> 05:47:46.317 rate and time. The key relationship to keep in mind is that the rate of travel is the 05:47:46.317 --> 05:47:51.559 distance traveled divided by the time it takes to travel again. For example, if you're driving 05:47:51.560 --> 05:47:57.888 60 miles an hour, that's your rate. And that's because you're going a distance of 60 miles 05:47:57.887 --> 05:48:02.750 in one hour. Sometimes it's handy to rewrite that relationship by multiplying both sides 05:48:02.750 --> 05:48:10.020 by T time. And that gives us that R times T is equal to D. In other words, distance 05:48:10.020 --> 05:48:15.727 is equal to rate times time. There's one more important principle to keep in mind. And that's 05:48:15.727 --> 05:48:23.590 the idea that rates add. For example, if you normally walk at three miles per hour, but 05:48:23.590 --> 05:48:29.709 you're walking on a moving sidewalk, that's going at a rate of two miles per hour, then 05:48:29.709 --> 05:48:34.919 your total speed of travel with respect to you know something stationary is going to 05:48:34.919 --> 05:48:43.179 be three plus two, or five miles per hour. All right, that is a formula as our one or 05:48:43.180 --> 05:48:51.207 the first rate per second rate is equal to the total rate. Let's use those two key ideas. 05:48:51.207 --> 05:48:59.000 distance equals rate times time, and rates add in the following problem. else's boat 05:48:59.000 --> 05:49:04.119 has a top speed of six miles per hour and still water. While traveling on a river at 05:49:04.119 --> 05:49:10.319 top speed. She went 10 miles upstream in the same amount of time she went 30 miles downstream, 05:49:10.319 --> 05:49:15.579 we're supposed to find the rate of the river currents. I'm going to organize the information 05:49:15.580 --> 05:49:17.568 in this problem into a chart. 05:49:17.567 --> 05:49:23.110 During the course of Elsa stay, there were two situations we need to keep in mind. For 05:49:23.110 --> 05:49:27.779 one period of time she was going upstream. And for another period of time she was going 05:49:27.779 --> 05:49:35.439 downstream. For each of those, I'm going to chart out the distance you traveled. The rate 05:49:35.439 --> 05:49:42.109 she went at and the time it took when she was going upstream. She went a total distance 05:49:42.110 --> 05:49:49.547 of 10 miles. When she was going downstream she went a longer distance of 30 miles. But 05:49:49.547 --> 05:49:55.207 the times to travel those two distances were the same. Since I don't know what that time 05:49:55.207 --> 05:50:02.930 was, I'll just give it a variable I'll call it T now Think about her rate of travel, the 05:50:02.930 --> 05:50:07.189 rate she traveled upstream was slower because she was going against the current and faster 05:50:07.189 --> 05:50:11.889 when she was going downstream with the current. We don't know what the speed of the current 05:50:11.889 --> 05:50:17.628 is, that's what we're trying to figure out. So maybe I'll give it a variable R. But we 05:50:17.628 --> 05:50:23.488 do know that in still water also can go six miles per hour. And when she's going downstream, 05:50:23.488 --> 05:50:29.260 since she's going with the direction of the current rates should add, and her rate downstream 05:50:29.259 --> 05:50:40.279 should be six plus R, that's her rate, and still water plus the rate of the current. 05:50:40.279 --> 05:50:46.707 On the other hand, when she's going upstream, then she's going against the current, so her 05:50:46.707 --> 05:50:53.149 rate of six miles per hour, we need to subtract the rate of the current from that. Now that 05:50:53.150 --> 05:50:57.250 we've charted out our information, we can turn it into equations using the fact that 05:50:57.250 --> 05:51:04.137 distance equals rate times time, we actually have two equations 10 equals six minus R times 05:51:04.137 --> 05:51:12.047 T, and 30 is equal to six plus R times T. Now that we've converted our situation into 05:51:12.047 --> 05:51:17.149 a system of equations, our next job is to solve the system of equations. In this example, 05:51:17.150 --> 05:51:23.580 I think the easiest way to proceed is to isolate t in each of the two equations. So in the 05:51:23.580 --> 05:51:28.200 first equation, I'll divide both sides by six minus r, and a second equation R divided 05:51:28.200 --> 05:51:36.760 by six plus R. That gives me 10 over six minus r equals T, and 30 over six plus r equals 05:51:36.759 --> 05:51:45.477 t. Now if I set my T variables equal to each other, I get, I get 10 over six minus r is 05:51:45.477 --> 05:51:52.989 equal to 30 over six plus R. I'm making progress because now I have a single equation, the 05:51:52.990 --> 05:51:58.670 single variable that I need to solve. Since the variable R is trapped in the denominator, 05:51:58.669 --> 05:52:03.887 I'm going to proceed by clearing the denominator. So I'm going to multiply both sides by the 05:52:03.887 --> 05:52:14.529 least common denominator, that is six minus r times six plus R. Once I cancel things out, 05:52:14.529 --> 05:52:23.689 I get that the six plus r times 10 is equal to 30 times six minus r, if I distribute, 05:52:23.689 --> 05:52:36.637 I'm going to get 60 plus xR equals 180 minus 30 ar, which I can now solve, let's see, that's 05:52:36.637 --> 05:52:43.909 going to be 40 r is equal to 120. So our, the speed of my current is going to be three 05:52:43.909 --> 05:52:51.099 miles per hour. This is all that the problem asked for the speed of the current. If I also 05:52:51.099 --> 05:52:57.579 wanted to solve for the other unknown time, I could do so by plugging in R into one of 05:52:57.580 --> 05:53:04.270 my equations and solving for T. In this video, I saw the distance rate and time problem by 05:53:04.270 --> 05:53:12.279 charting out my information for the two situations and my problem using the fact that rates add 05:53:12.279 --> 05:53:17.930 to fill in some of my boxes, and then using the formula distance equals rate times time 05:53:17.930 --> 05:53:24.080 to build a system of equations. In this video, I'll do a standard mixture problem in which 05:53:24.080 --> 05:53:30.340 we have to figure out what quantity of two solutions to mix together. household bleach 05:53:30.340 --> 05:53:39.387 contains 6% sodium hypochlorite. The other 94% is water. How much household bleach should 05:53:39.387 --> 05:53:50.957 be combined with 70 liters of a weaker 1% hypochlorite solution in order to form a solution, 05:53:50.957 --> 05:53:59.349 that's 2.5% sodium hypochlorite. I want to turn this problem into a system of equations. 05:53:59.349 --> 05:54:05.977 So I'm asking myself what quantities are going to be equal to each other? Well, the total 05:54:05.977 --> 05:54:14.829 amount of sodium hypochlorite that has symbol and a co o before mixing, 05:54:14.830 --> 05:54:23.270 it should equal the total amount of sodium hypochlorite after mixing. Also, the total 05:54:23.270 --> 05:54:30.378 amount of water before mixing should equal the total amount of water after. And finally, 05:54:30.378 --> 05:54:37.790 there's just the total amount of solution. In my two jugs, sodium hypochlorite together 05:54:37.790 --> 05:54:47.290 with water should equal the total amount of solution after that gives me a hint for what 05:54:47.290 --> 05:54:51.909 I'm looking for. But before I start reading out equations, I find it very helpful to chart 05:54:51.909 --> 05:55:05.387 out my quantities. So I've got the 6% solution. The household leech, I've got the 1% solution. 05:55:05.387 --> 05:55:16.128 And I've got my Desired Ending 2.5% solution. Now on each of those solutions, I've got a 05:55:16.128 --> 05:55:26.580 certain volume of sodium hypochlorite. I've also got a volume of water. And I've got a 05:55:26.580 --> 05:55:37.200 total volume of solution. Let me see which of these boxes I can actually fill in, I know 05:55:37.200 --> 05:55:45.640 that I'm adding 70 liters of the 1% solution. So I can put a 70 in the total volume of solution 05:55:45.640 --> 05:55:53.509 here. I don't know what volume of the household bleach, I want to add, that's what I'm trying 05:55:53.509 --> 05:56:02.949 to find out. So I'm going to just call that volume x. Now since my 2.5% solution is made 05:56:02.950 --> 05:56:08.409 by combining my other two solutions, I know its volume is going to be the sum of these 05:56:08.409 --> 05:56:16.680 two volumes, so I'll write 70 plus x in this box. Now, the 6% solution means that whatever 05:56:16.680 --> 05:56:22.580 the volume of solution is, 6% of that is the sodium hypochlorite. So the volume of the 05:56:22.580 --> 05:56:31.070 sodium hypochlorite is going to be 0.06 times x, the volume of water and that solution is 05:56:31.069 --> 05:56:40.000 whatever's left, so that's going to be x minus 0.06x, or point nine four times X with the 05:56:40.000 --> 05:56:46.988 following the same reasoning for the 1% solution 1% of the 70 liters is a sodium hypochlorite. 05:56:46.988 --> 05:56:56.530 So that's going to be 0.01 times 70. Or point seven, the volume of water and that solution 05:56:56.529 --> 05:57:05.699 is going to be 99% or point nine, nine times seven day. That works out to 69.3. Finally, 05:57:05.700 --> 05:57:15.317 for the 2.5% solution, the volume of the sodium hypochlorite is going to be 0.0 to five times 05:57:15.317 --> 05:57:22.599 the volume of solution 70 plus x and the volume of water is going to be the remainder. So 05:57:22.599 --> 05:57:32.919 that's 0.975 times 70 plus x. Now I've already used the fact that the volume of solutions 05:57:32.919 --> 05:57:39.359 before added up is the volume of solution after in writing a 70 plus x in this box. 05:57:39.360 --> 05:57:43.830 But I haven't yet used the fact that the volume of the sodium hypochlorite is preserved before 05:57:43.830 --> 05:57:53.040 and after. So I can write that down as an equation. So that means 0.06x plus 0.7 is 05:57:53.040 --> 05:58:01.270 equal to 0.025 times 70 plus x. Now I've got an equation with a variable. I'll try to solve 05:58:01.270 --> 05:58:06.850 it. Since I don't like all these decimals, I'm going to multiply both sides of my equation 05:58:06.849 --> 05:58:15.057 by let's see, 1000 should get rid of all the decimals. After distributing, I get 60x plus 05:58:15.058 --> 05:58:27.090 700 equals 25 times 70 plus x. Distributing some more, I get 60x plus 700 is equal to 05:58:27.090 --> 05:58:42.409 1750 plus 25x. So let's see 60 minus 25 is 35x is equal to 1050. Which gives me that 05:58:42.409 --> 05:58:49.139 x equals 30 liters of the household bleach. 05:58:49.139 --> 05:58:54.610 Notice that I never actually had to use the fact that the water quantity of water before 05:58:54.610 --> 05:59:00.659 mixing is equal to the quantity of water after that is I never use the information in this 05:59:00.659 --> 05:59:08.450 column. In fact, that information is redundant. Once I know that the quantities of sodium 05:59:08.450 --> 05:59:15.520 hypochlorite add up, and the total volume of solutions add up. The fact that that volume 05:59:15.520 --> 05:59:21.689 of waters add up is just redundant information. The techniques to use to solve this equation 05:59:21.689 --> 05:59:28.039 involving solutions can be used to solve many many other equations involving mixtures of 05:59:28.040 --> 05:59:36.977 items. My favorite method is to first make a chart involving the types of mixtures and 05:59:36.977 --> 05:59:43.930 the types of items in your mixture. Fill in as many boxes size I can and then use the 05:59:43.930 --> 05:59:53.159 fact that the quantities add. This video is about rational functions and their graphs. 05:59:53.159 --> 05:59:59.000 Recall that a rational function is a function that can be written as a ratio or quotient 05:59:59.000 --> 06:00:07.308 of two power. No Here's an example. The simpler function, f of x equals one over x is also 06:00:07.308 --> 06:00:14.520 considered a rational function, you can think of one and x as very simple polynomials. The 06:00:14.520 --> 06:00:20.110 graph of this rational function is shown here. This graph looks different from the graph 06:00:20.110 --> 06:00:27.290 of a polynomial. For one thing, its end behavior is different. The end behavior of a function 06:00:27.290 --> 06:00:33.030 is the way the graph looks when x goes through really large positive, or really large negative 06:00:33.029 --> 06:00:38.840 numbers, we've seen that the end behavior of a polynomial always looks like one of these 06:00:38.840 --> 06:00:44.700 cases. That is why marches off to infinity or maybe negative infinity, as x gets really 06:00:44.700 --> 06:00:50.200 big or really negative. But this rational function has a different type of end behavior. 06:00:50.200 --> 06:00:54.600 Notice, as x gets really big, the y values are leveling off 06:00:54.599 --> 06:01:00.779 at about a y value of three. And similarly, as x values get really negative, our graph 06:01:00.779 --> 06:01:07.849 is leveling off near the line y equals three, I'll draw that line, y equals three on my 06:01:07.849 --> 06:01:16.119 graph, that line is called a horizontal asymptote. A horizontal asymptote is a horizontal line 06:01:16.119 --> 06:01:21.529 that our graph gets closer and closer to as x goes to infinity, or as X goes to negative 06:01:21.529 --> 06:01:26.750 infinity, or both. There's something else that's different about this graph from a polynomial 06:01:26.750 --> 06:01:32.599 graph, look at what happens as x gets close to negative five. As we approach negative 06:01:32.599 --> 06:01:37.717 five with x values on the right, our Y values are going down towards negative infinity. 06:01:37.718 --> 06:01:42.690 And as we approach the x value of negative five from the left, our Y values are going 06:01:42.689 --> 06:01:49.340 up towards positive infinity. We say that this graph has a vertical asymptote at x equals 06:01:49.340 --> 06:01:56.439 negative five. A vertical asymptote is a vertical line that the graph gets closer and closer 06:01:56.439 --> 06:02:02.567 to. Finally, there's something really weird going on at x equals two, there's a little 06:02:02.567 --> 06:02:10.000 open circle there, like the value at x equals two is dug out. That's called a hole. A hole 06:02:10.000 --> 06:02:16.718 is a place along the curve of the graph where the function doesn't exist. Now that we've 06:02:16.718 --> 06:02:19.600 identified some of the features of our rational functions graph, 06:02:19.599 --> 06:02:24.627 I want to look back at the equation and see how we could have predicted those features 06:02:24.628 --> 06:02:31.040 just by looking at the equation. To find horizontal asymptotes, we need to look at what our function 06:02:31.040 --> 06:02:36.580 is doing when x goes through really big positive or really big negative numbers. Looking at 06:02:36.580 --> 06:02:42.990 our equation for our function, the numerator is going to be dominated by the 3x squared 06:02:42.990 --> 06:02:47.659 term when x is really big, right, because three times x squared is going to be absolutely 06:02:47.659 --> 06:02:53.619 enormous compared to this negative 12. If x is a big positive or negative number, in 06:02:53.619 --> 06:02:59.610 the denominator, the denominator will be dominated by the x squared term. Again, if x is a really 06:02:59.610 --> 06:03:04.920 big positive or negative number, like a million, a million squared will be much, much bigger 06:03:04.919 --> 06:03:10.797 than three times a million or negative 10. For that reason, to find the end behavior, 06:03:10.797 --> 06:03:18.387 or the horizontal asymptote, for our function, we just need to look at the term on the numerator 06:03:18.387 --> 06:03:22.637 and the term on the denominator that have the highest exponent, those are the ones that 06:03:22.637 --> 06:03:29.468 dominate the expression in size. So as x gets really big, our functions y values are going 06:03:29.468 --> 06:03:37.220 to be approximately 3x squared over x squared, which is three. That's why we have a horizontal 06:03:37.220 --> 06:03:45.430 asymptote at y equals three. Now our vertical asymptotes, those tend to occur where our 06:03:45.430 --> 06:03:51.308 denominator of our function is zero. That's because the function doesn't exist when our 06:03:51.308 --> 06:03:56.150 denominator is zero. And when we get close to that place where our denominator is zero, 06:03:56.150 --> 06:04:01.680 we're going to be dividing by tiny, tiny numbers, which will make our Y values really big in 06:04:01.680 --> 06:04:06.920 magnitude. So to check where our denominators zero, let's factor our function. In fact, 06:04:06.919 --> 06:04:11.909 I'm going to go ahead and factor the numerator and the denominator. So the numerator factors, 06:04:11.909 --> 06:04:18.468 let's see, pull out the three, I get x squared minus four, factor in the denominator, that 06:04:18.468 --> 06:04:25.040 factors into X plus five times x minus two, I can factor a little the numerator a little 06:04:25.040 --> 06:04:34.950 further, that's three times x minus two times x plus two over x plus 5x minus two. Now, 06:04:34.950 --> 06:04:41.319 when x is equal to negative five, my denominator will be zero, but my numerator will not be 06:04:41.319 --> 06:04:49.398 zero. That's what gives me the vertical asymptote at x equals negative five. Notice that when 06:04:49.398 --> 06:04:57.292 x equals two, the denominator is zero, but the numerator is also zero. In fact, if I 06:04:57.292 --> 06:05:01.909 cancelled the x minus two factor from the numerator, and in nominator, I get a simplified 06:05:01.909 --> 06:05:11.279 form for my function that agrees with my original function as long as x is not equal to two. 06:05:11.279 --> 06:05:16.760 That's because when x equals two, the simplified function exists, but the original function 06:05:16.760 --> 06:05:23.637 does not, it's zero over zero, it's undefined. But for every other x value, including x values 06:05:23.637 --> 06:05:29.829 near x equals to our original functions just the same as this function. And that's why 06:05:29.830 --> 06:05:36.468 our function only has a vertical asymptote at x equals negative five, not one at x equals 06:05:36.468 --> 06:05:41.069 two, because the x minus two factor is no longer in the function after simplifying, 06:05:41.069 --> 06:05:45.779 it does have a hole at x equals two, because the original function is not defined there, 06:05:45.779 --> 06:05:52.189 even though the simplified version is if we want to find the y value of our hole, we can 06:05:52.189 --> 06:05:59.599 just plug in x equals two into our simplified version of our function, that gives a y value 06:05:59.599 --> 06:06:06.769 of three times two plus two over two plus seven, or 12 ninths, which simplifies to four 06:06:06.770 --> 06:06:15.477 thirds. So our whole is that to four thirds. Now that we've been through one example in 06:06:15.477 --> 06:06:22.520 detail, let's summarize our findings. We find the vertical asymptotes and the holes by looking 06:06:22.520 --> 06:06:29.477 where the denominator is zero. The holes happen where the denominator and numerator are both 06:06:29.477 --> 06:06:35.930 zero and those factors cancel out. The vertical asymptotes are all other x values where the 06:06:35.930 --> 06:06:42.010 denominator is zero, we find the horizontal asymptotes. By considering the highest power 06:06:42.009 --> 06:06:47.477 term on the numerator and the denominator, I'll explain this process in more detail in 06:06:47.477 --> 06:06:55.567 three examples. In the first example, if we circle the highest power terms, that simplifies 06:06:55.567 --> 06:07:03.648 to 5x over 3x squared, which is five over 3x. As x gets really big, the denominator 06:07:03.648 --> 06:07:09.958 is going to be huge. So I'm going to be dividing five by a huge, huge number, that's going 06:07:09.957 --> 06:07:15.147 to be going very close to zero. And therefore we have a horizontal asymptote 06:07:15.148 --> 06:07:24.398 at y equals zero. In the second example, the highest power terms, 2x cubed, over 3x cubed 06:07:24.398 --> 06:07:30.200 simplifies to two thirds. So as x gets really big, we're going to be heading towards two 06:07:30.200 --> 06:07:36.950 thirds, and we have a horizontal asymptote at y equals two thirds. In the third example, 06:07:36.950 --> 06:07:45.270 the highest power terms, x squared over 2x simplifies to x over two. As x gets really 06:07:45.270 --> 06:07:51.887 big, x over two is getting really big. And therefore, we don't have a horizontal asymptote 06:07:51.887 --> 06:07:59.279 at all. This is going to infinity, when x gets through go through big positive numbers, 06:07:59.279 --> 06:08:07.237 and is going to negative infinity when x goes through a big negative numbers. So in this 06:08:07.238 --> 06:08:12.840 case, the end behavior is kind of like that of a polynomial, and there's no horizontal 06:08:12.840 --> 06:08:18.887 asymptote. In general, when the degree of the numerator is smaller than the degree of 06:08:18.887 --> 06:08:23.520 the denominator, we're in this first case where the denominator gets really big compared 06:08:23.520 --> 06:08:29.250 to the numerator and we go to zero. In the second case, where the degree of the numerator 06:08:29.250 --> 06:08:34.297 and the degree of the dominant are equal, things cancel out, and so we get a horizontal 06:08:34.297 --> 06:08:42.750 asymptote at the y value, that's equal to the ratio of the leading coefficients. Finally, 06:08:42.750 --> 06:08:46.968 in the third case, when the degree of the numerator is bigger than the degree of the 06:08:46.968 --> 06:08:52.290 denominator, then the numerator is getting really big compared to the denominator, so 06:08:52.290 --> 06:08:57.840 we end up with no horizontal asymptote. Final Finally, let's apply all these observations 06:08:57.840 --> 06:09:03.379 to one more example. Please pause the video and take a moment to find the vertical asymptotes, 06:09:03.379 --> 06:09:09.817 horizontal asymptotes and holes for this rational function. To find the vertical asymptotes 06:09:09.817 --> 06:09:15.547 and holes, we need to look at where the denominator is zero. In fact, it's going to be handy to 06:09:15.547 --> 06:09:20.289 factor both the numerator and the denominator. Since there if there are any common factors, 06:09:20.290 --> 06:09:25.128 we might have a whole instead of a vertical asymptote. The numerator is pretty easy to 06:09:25.128 --> 06:09:31.887 factor. Let's see that's 3x times x plus one for the denominator or first factor out an 06:09:31.887 --> 06:09:40.029 x. And then I'll factor some more using a guess and check method. I know that I'll need 06:09:40.029 --> 06:09:47.967 a 2x and an X to multiply together to the 2x squared and I'll need a three and a minus 06:09:47.968 --> 06:09:56.030 one or alpha minus three and a one. Let's see if that works. If I multiply out 2x minus 06:09:56.029 --> 06:10:02.860 one times x plus three that does get me back to x squared plus 5x minus three, so that 06:10:02.861 --> 06:10:08.590 checks out. Now I noticed that I have a common factor of x in both the numerator and the 06:10:08.590 --> 06:10:14.270 denominator. So that's telling me I'm going to have a hole at x equals zero. In fact, 06:10:14.270 --> 06:10:21.889 I could rewrite my rational function by cancelling out that common factor. And that's equivalent, 06:10:21.889 --> 06:10:28.290 as long as x is not equal to zero. So the y value of my whole is what I get when I plug 06:10:28.290 --> 06:10:35.628 zero into my simplified version, that would be three times zero plus one over two times 06:10:35.628 --> 06:10:42.797 zero minus one times zero plus three, which is three over negative three or minus one. 06:10:42.797 --> 06:10:49.619 So my whole is at zero minus one. Now all the remaining places in my denominator that 06:10:49.619 --> 06:10:55.239 make my denominator zero will get me vertical asymptotes. So I'll have a vertical asymptote, 06:10:55.240 --> 06:11:04.780 when 2x minus one times x plus three equals zero, that is, when 2x minus one is zero, 06:11:04.779 --> 06:11:13.840 or x plus three is zero. In other words, when x is one half, or x equals negative three. 06:11:13.840 --> 06:11:20.299 Finally, to find my horizontal asymptotes, I just need to consider the highest power 06:11:20.299 --> 06:11:28.759 term in the numerator and the denominator. That simplifies to three over 2x, which is 06:11:28.759 --> 06:11:35.359 bottom heavy, right? When x gets really big, this expression is going to zero. And that 06:11:35.360 --> 06:11:41.718 means that we have a horizontal asymptote at y equals zero. So we found the major features 06:11:41.718 --> 06:11:48.150 of our graph, the whole, the vertical asymptotes and the horizontal asymptotes. Together, this 06:11:48.150 --> 06:11:54.458 would give us a framework for what the graph of our function looks like. horizontal asymptote 06:11:54.457 --> 06:12:02.599 at y equals zero, vertical asymptotes at x equals one half, and x equals minus three 06:12:02.599 --> 06:12:09.909 at a hole at the point zero minus one. plotting a few more points, or using a graphing calculator 06:12:09.909 --> 06:12:17.950 of graphing program, we can see that our actual function will look something like this. 06:12:17.950 --> 06:12:24.159 Notice that the x intercept when x is negative one, corresponds to where the numerator of 06:12:24.159 --> 06:12:29.529 our rational function or reduced rational function is equal to zero. That's because 06:12:29.529 --> 06:12:33.949 a zero on the numerator that doesn't make the denominator zero makes the whole function 06:12:33.950 --> 06:12:40.000 zero. And an X intercept is where the y value of the whole function is zero. In this video, 06:12:40.000 --> 06:12:44.797 we learned how to find horizontal asymptotes have rational functions. By looking at the 06:12:44.797 --> 06:12:50.779 highest power terms, we learned to find the vertical asymptotes and holes. By looking 06:12:50.779 --> 06:12:56.307 at the factored version of the functions. The holes correspond to the x values that 06:12:56.308 --> 06:13:03.317 make the numerator and denominator zero, his corresponding factors cancel. The vertical 06:13:03.317 --> 06:13:09.099 asymptotes correspond to the x values that make the denominator zero, even after factoring 06:13:09.099 --> 06:13:13.957 any any common and in common factors in the numerator denominator. 06:13:13.957 --> 06:13:19.750 This video is about combining functions by adding them subtracting them multiplying and 06:13:19.750 --> 06:13:26.270 dividing them. Suppose we have two functions, f of x equals x plus one and g of x equals 06:13:26.270 --> 06:13:34.218 x squared. One way to combine them is by adding them together. This notation, f plus g of 06:13:34.218 --> 06:13:43.909 x means the function defined by taking f of x and adding it to g of x. So for our functions, 06:13:43.909 --> 06:13:50.957 that means we take x plus one and add x squared, I can rearrange that as the function x squared 06:13:50.957 --> 06:14:01.899 plus x plus one. So f plus g evaluated on x means x squared plus x plus one. And if 06:14:01.900 --> 06:14:08.810 I wanted to evaluate f plus g, on the number two, that would be two squared plus two plus 06:14:08.810 --> 06:14:18.218 one, or seven. Similarly, the notation f minus g of x means the function we get by taking 06:14:18.218 --> 06:14:25.490 f of x and subtracting g of x. So that would be x plus one minus x squared. And if I wanted 06:14:25.490 --> 06:14:35.390 to take f minus g evaluated at one, that would be one plus one minus one squared, or one, 06:14:35.389 --> 06:14:44.599 the notation F dot g of x, which is sometimes also written just as f g of x. That means 06:14:44.599 --> 06:14:52.969 we take f of x times g of x. In other words, x plus one times x squared, which could be 06:14:52.970 --> 06:15:02.888 simplified as x cubed plus x squared. The notation f divided by g of x means I take 06:15:02.887 --> 06:15:11.919 f of x and divided by g of x. So that would be x plus one divided by x squared. In this 06:15:11.919 --> 06:15:19.717 figure, the blue graph represents h of x. And the red graph represents the function 06:15:19.718 --> 06:15:26.690 p of x, we're asked to find h minus p of zero. 06:15:26.689 --> 06:15:32.259 We don't have any equations to work with, but that's okay. We know that for any x, h 06:15:32.259 --> 06:15:40.877 minus p of x is defined as h of x minus p of x. So for x equals zero, H minus p of zero 06:15:40.878 --> 06:15:48.968 is going to be h of zero minus p of zero. Using the graph, we can find h of zero by 06:15:48.968 --> 06:15:55.830 finding the value of zero on the x axis, and finding the corresponding y value for the 06:15:55.830 --> 06:16:04.120 function h of x. So that's about 1.8. Now P of zero, we can find similarly by looking 06:16:04.119 --> 06:16:10.090 for zero on the x axis, and finding the corresponding y value for the function p of x, and that's 06:16:10.090 --> 06:16:18.939 a y value of one 1.8 minus one is 0.8. So that's our approximate value for H minus p 06:16:18.939 --> 06:16:28.699 of zero. If we want to find P times h of negative three, again, we can rewrite that as P of 06:16:28.700 --> 06:16:34.899 negative three times h of negative three. And using the graphs, we see that for an x 06:16:34.899 --> 06:16:45.190 value of negative three, the y value for P is two. And the x value of negative three 06:16:45.189 --> 06:16:50.180 corresponds to a y value of negative two for H. 06:16:50.180 --> 06:16:55.610 Two times negative two is negative four. So that's our value for P times h of negative 06:16:55.610 --> 06:16:56.610 three. 06:16:56.610 --> 06:17:03.878 In this video, we saw how to add two functions, subtract two functions, multiply two functions, 06:17:03.878 --> 06:17:13.290 and divide two functions in the following way. When you compose two functions, you apply 06:17:13.290 --> 06:17:19.940 the first function, and then you apply the second function to the output of the first 06:17:19.939 --> 06:17:27.349 function. For example, the first function might compute population size from time in 06:17:27.349 --> 06:17:35.807 years. So its input would be time in years, since a certain date, as output would be number 06:17:35.808 --> 06:17:45.620 of people in the population. The second function g, might compute health care costs as a function 06:17:45.619 --> 06:17:53.919 of population size. So it will take population size as input, and its output will be healthcare 06:17:53.919 --> 06:18:00.189 costs. If you put these functions together, that is compose them, then you'll go all the 06:18:00.189 --> 06:18:07.770 way from time in years to healthcare costs. This is your composition, g composed with 06:18:07.770 --> 06:18:16.409 F. The composition of two functions, written g with a little circle, f of x is defined 06:18:16.409 --> 06:18:26.919 as follows. g composed with f of x is G evaluated on f of x, we can think of it schematically 06:18:26.919 --> 06:18:36.009 and diagram f x on a number x and produces a number f of x, then g takes that output 06:18:36.009 --> 06:18:45.137 f of x and produces a new number, g of f of x. Our composition of functions g composed 06:18:45.137 --> 06:18:51.950 with F is the function that goes all the way from X to g of f of x. Let's work out some 06:18:51.950 --> 06:18:58.920 examples where our functions are defined by tables of values. If we want to find g composed 06:18:58.919 --> 06:19:08.189 with F of four, by definition, this means g of f of four. To evaluate this expression, 06:19:08.189 --> 06:19:15.029 we always work from the inside out. So we start with the x value of four, and we find 06:19:15.029 --> 06:19:23.739 f of four, using the table of values for f of x, when x equals four, f of x is seven, 06:19:23.740 --> 06:19:33.128 so we can replace F of four with the number seven. Now we need to evaluate g of seven, 06:19:33.128 --> 06:19:39.420 seven becomes our new x value in our table of values for G, the x value of seven corresponds 06:19:39.419 --> 06:19:47.949 to the G of X value of 10. So g of seven is equal to 10. We found that g composed with 06:19:47.950 --> 06:19:56.387 F of four is equal to 10. If instead we want to find f composed with g of four, well, we 06:19:56.387 --> 06:20:04.547 can rewrite that as f of g of four Again work from the inside out. Now we're trying to find 06:20:04.547 --> 06:20:10.250 g of four. So four is our x value. And we use our table of values for G to see that 06:20:10.250 --> 06:20:20.430 g of four is one. So we replaced you a four by one. And now we need to evaluate f of one. 06:20:20.430 --> 06:20:29.817 Using our table for F values, f of one is eight. Notice that when we've computed g of 06:20:29.817 --> 06:20:36.907 f of four, we got a different answer than when we computed f of g of four. And in general, 06:20:36.907 --> 06:20:43.509 g composed with F is not the same thing as f composed with g. Please pause the video 06:20:43.509 --> 06:20:49.808 and take a moment to compute the next two examples. We can replace f composed with F 06:20:49.809 --> 06:20:56.708 of two by the equivalent expression, f of f of two. Working from the inside out, we 06:20:56.707 --> 06:21:06.127 know that f of two is three, and f of three is six. If we want to find f composed with 06:21:06.128 --> 06:21:17.010 g of six, rewrite that as f of g of six, using the table for g, g of six is eight. But F 06:21:17.009 --> 06:21:25.109 of eight, eight is not on the table as an x value for the for the f function. And so 06:21:25.110 --> 06:21:35.148 there is no F of eight, this does not exist, we can say that six is not in the domain, 06:21:35.148 --> 06:21:42.970 for F composed with g. Even though it was in the domain of g, we couldn't follow all 06:21:42.970 --> 06:21:49.780 the way through and get a value for F composed with g of six. Next, let's turn our attention 06:21:49.779 --> 06:21:54.399 to the composition of functions that are given by equations. 06:21:54.400 --> 06:22:03.110 p of x is x squared plus x and q of x as negative 2x. We want to find q composed with P of one. 06:22:03.110 --> 06:22:14.047 As usual, I can rewrite this as Q of P of one and work from the inside out. P of one 06:22:14.047 --> 06:22:21.567 is one squared plus one, so that's two. So this is the same thing as Q of two. But Q 06:22:21.567 --> 06:22:28.009 of two is negative two times two or negative four. So this evaluates to negative four. 06:22:28.009 --> 06:22:35.237 In this next example, we want to find q composed with P of some arbitrary x, or rewrite it 06:22:35.238 --> 06:22:44.370 as usual as Q of p of x and work from the inside out. Well, p of x, we know the formula 06:22:44.369 --> 06:22:52.047 for that. That's x squared plus x. So I can replace my P of x with that expression. Now, 06:22:52.047 --> 06:22:59.919 I'm stuck with evaluating q on x squared plus x. Well, Q of anything is negative two times 06:22:59.919 --> 06:23:08.239 that thing. So q of x squared plus x is going to be negative two times the quantity x squared 06:23:08.240 --> 06:23:13.808 plus x, what I've done is I've substituted in the whole expression x squared plus x, 06:23:13.808 --> 06:23:20.010 where I saw the X in this formula for q of x, it's important to use the parentheses here. 06:23:20.009 --> 06:23:24.297 So that will be multiplying negative two by the whole expression and not just by the first 06:23:24.297 --> 06:23:32.147 piece, I can simplify this a bit as negative 2x squared minus 2x. And that's my expression 06:23:32.148 --> 06:23:40.478 for Q composed with p of x. Notice that if I wanted to compute q composed with P of one, 06:23:40.477 --> 06:23:45.309 which I already did in the first problem, I could just use this expression now, negative 06:23:45.310 --> 06:23:52.740 two times one squared minus two and I get negative four, just like I did before. Let's 06:23:52.740 --> 06:24:00.340 try another one. Let's try p composed with q of x. First I read write this P of q of 06:24:00.340 --> 06:24:07.009 x. Working from the inside out, I can replace q of x with negative 2x. So I need to compute 06:24:07.009 --> 06:24:15.419 P of negative 2x. Here's my formula for P. to compute P of this expression, I need to 06:24:15.419 --> 06:24:23.657 plug in this expression everywhere I see an x in the formula for P. So that means negative 06:24:23.657 --> 06:24:29.819 2x squared plus negative 2x. Again, being careful to use parentheses to make sure I 06:24:29.819 --> 06:24:40.457 plug in the entire expression in forex. let me simplify. This is 4x squared minus 2x. 06:24:40.457 --> 06:24:50.407 Notice that I got different expressions for Q of p of x, and for P of q of x. Once again, 06:24:50.407 --> 06:24:57.720 we see that q composed with P is not necessarily equal to P composed with Q. Please pause the 06:24:57.720 --> 06:25:06.850 video and try this last example yourself. rewriting. And working from the inside out, 06:25:06.849 --> 06:25:13.750 we're going to replace p of x with its expression x squared plus x. And then we need to evaluate 06:25:13.750 --> 06:25:22.090 p on x squared plus x. That means we plug in x squared plus x, everywhere we see an 06:25:22.090 --> 06:25:29.657 x in this formula, so that's x squared plus x quantity squared plus x squared plus x. 06:25:29.657 --> 06:25:36.809 Once again, I can simplify by distributing out, that gives me x to the fourth plus 2x 06:25:36.810 --> 06:25:45.020 cubed plus x squared plus x squared plus x, or x to the fourth plus 2x cubed plus 2x squared 06:25:45.020 --> 06:25:52.189 plus x. In this last set of examples, we're asked to go backwards, we're given a formula 06:25:52.189 --> 06:25:57.919 for a function of h of x. But we're supposed to rewrite h of x as a composition of two 06:25:57.919 --> 06:26:04.359 functions, F and G. Let's think for a minute, which of these two functions gets applied 06:26:04.360 --> 06:26:13.619 first, f composed with g of x, let's see, that means f of g of x. And since we evaluate 06:26:13.619 --> 06:26:21.250 these expressions from the inside out, we must be applying g first, and then F. In order 06:26:21.250 --> 06:26:27.779 to figure out what what f and g could be, I like to draw a box around some thing inside 06:26:27.779 --> 06:26:32.779 my expression for H, so I'm going to draw a box around x squared plus seven, then whatever's 06:26:32.779 --> 06:26:38.529 inside the box, that'll be my function, g of x, the first function that gets applied, 06:26:38.529 --> 06:26:43.849 whatever happens to the box, in this case, taking the square root sign, that becomes 06:26:43.849 --> 06:26:46.869 my outside function, my second function f. 06:26:46.869 --> 06:26:54.750 So here, we're gonna say g of x is equal to x squared plus seven, and f of x is equal 06:26:54.750 --> 06:27:00.939 to the square root of x, let's just check and make sure that this works. So I need to 06:27:00.939 --> 06:27:08.069 check that when I take the composition, f composed with g, I need to get the same thing 06:27:08.069 --> 06:27:17.029 as my original h. So let's see, if I do f composed with g of x, well, by definition, 06:27:17.029 --> 06:27:23.397 that's f of g of x, working from the inside out, I can replace g of x with its formula 06:27:23.398 --> 06:27:30.978 x squared plus seven. So I need to evaluate f of x squared plus seven. That means I plug 06:27:30.977 --> 06:27:37.500 in x squared plus seven, into the formula for for F. So that becomes the square root 06:27:37.500 --> 06:27:44.128 of x squared plus seven to the it works because it matches my original equation. So we found 06:27:44.128 --> 06:27:48.180 a correct answer a correct way of breaking h down as a composition of two functions. 06:27:48.180 --> 06:27:54.680 But I do want to point out, this is not the only correct answer. I'll write down my formula 06:27:54.680 --> 06:27:59.738 for H of X again, and this time, I'll put the box in a different place, I'll just box 06:27:59.738 --> 06:28:08.128 the x squared. If I did that, then my inside function of my first function, g of x would 06:28:08.128 --> 06:28:17.450 be x squared. And my second function is what happens to the box. So my f of x is what happens 06:28:17.450 --> 06:28:23.898 to the box, and the box gets added seven to it, and taking the square root. So in other 06:28:23.898 --> 06:28:32.260 words, f of x is going to be the square root of x plus seven. Again, I can check that this 06:28:32.259 --> 06:28:39.727 works. If I do f composed with g of x, that's f of g of x. So now g of x is x squared. So 06:28:39.727 --> 06:28:46.390 I'm taking f of x squared. When I plug in x squared for x, I do in fact, get the square 06:28:46.390 --> 06:28:53.207 root of x squared plus seven. So this is that alternative, correct solution. In this video, 06:28:53.207 --> 06:29:00.590 we learn to evaluate the composition of functions. by rewriting it and working from the inside 06:29:00.590 --> 06:29:10.170 out. We also learn to break apart a complicated function into a composition of two functions 06:29:10.169 --> 06:29:16.339 by boxing one piece of the function and letting the first function applied in the composition. 06:29:16.340 --> 06:29:21.270 Let that be the inside of the box, and the second function applied in the composition 06:29:21.270 --> 06:29:33.540 be whatever happens to the box. 06:29:33.540 --> 06:29:38.580 The inverse of a function undoes what the function does. So the inverse of tying your 06:29:38.580 --> 06:29:46.750 shoes would be to untie them. And the inverse of the function that adds two to a number 06:29:46.750 --> 06:29:53.720 would be the function that subtracts two from a number. This video introduces inverses and 06:29:53.720 --> 06:30:00.378 their properties. Suppose f of x is a function defined by this chart. In other words, Have 06:30:00.378 --> 06:30:08.790 two is three, f of three is five, f of four is six, and f of five is one, the inverse 06:30:08.790 --> 06:30:17.817 function for F written f superscript. Negative 1x undoes what f does. Since f takes two to 06:30:17.817 --> 06:30:26.157 three, F inverse takes three, back to two. So we write this f superscript, negative one 06:30:26.157 --> 06:30:38.430 of three is to. Similarly, since f takes three to five, F inverse takes five to three. And 06:30:38.430 --> 06:30:46.860 since f takes four to six, f inverse of six is four. And since f takes five to one, f 06:30:46.860 --> 06:30:54.128 inverse of one is five. I'll use these numbers to fill in the chart. Notice that the chart 06:30:54.128 --> 06:31:00.450 of values when y equals f of x and the chart of values when y equals f inverse of x are 06:31:00.450 --> 06:31:07.208 closely related. They share the same numbers, but the x values for f of x correspond to 06:31:07.207 --> 06:31:14.647 the y values for f inverse of x, and the y values for f of x correspond to the x values 06:31:14.648 --> 06:31:21.138 for f inverse of x. That leads us to the first key fact inverse functions reverse the roles 06:31:21.137 --> 06:31:29.439 of y and x. I'm going to plot the points for y equals f of x in blue. Next, I'll plot the 06:31:29.439 --> 06:31:35.849 points for y equals f inverse of x in red. Pause the video for a moment and see what 06:31:35.849 --> 06:31:40.349 kind of symmetry you observe in this graph. How are the blue points related to the red 06:31:40.349 --> 06:31:46.737 points, you might have noticed that the blue points and the red points are mirror images 06:31:46.738 --> 06:31:55.250 over the mirror line, y equals x. So our second key fact is that the graph of y equals f inverse 06:31:55.250 --> 06:31:58.047 of x can be obtained from the graph of y equals f 06:31:58.047 --> 06:31:59.047 of x 06:31:59.047 --> 06:32:06.727 by reflecting over the line y equals x. This makes sense, because inverses, reverse the 06:32:06.727 --> 06:32:15.770 roles of war annex. In the same example, let's compute f inverse of f of two, this open circle 06:32:15.770 --> 06:32:23.878 means composition. In other words, we're computing f inverse of f of two, we compute this from 06:32:23.878 --> 06:32:33.350 the inside out. So that's f inverse of three. Since F of two is three, and f inverse of 06:32:33.349 --> 06:32:44.717 three, we see as to similarly, we can compute f of f inverse of three. And that means we 06:32:44.718 --> 06:32:52.690 take f of f inverse of three. Since f inverse of three is two, that's the same thing as 06:32:52.689 --> 06:33:02.039 computing F of two, which is three. Please pause the video for a moment and compute these 06:33:02.040 --> 06:33:11.148 other compositions. You should have found that in every case, if you take f inverse 06:33:11.148 --> 06:33:16.290 of f of a number, you get back to the very same number you started with. And similarly, 06:33:16.290 --> 06:33:20.738 if you take f of f inverse of any number, you get back to the same number you started 06:33:20.738 --> 06:33:29.909 with. So in general, f inverse of f of x is equal to x, and f of f inverse of x is also 06:33:29.909 --> 06:33:37.430 equal to x. This is the mathematical way of saying that F and n f inverse undo each other. 06:33:37.430 --> 06:33:43.229 Let's look at a different example. Suppose that f of x is x cubed. Pause the video for 06:33:43.229 --> 06:33:48.409 a moment, and guess what the inverse of f should be. Remember, F inverse undoes the 06:33:48.409 --> 06:33:57.180 work that F does. You might have guessed that f inverse of x is going to be the cube root 06:33:57.180 --> 06:34:05.680 function, we can check that this is true by looking at f of f inverse of x, that's F of 06:34:05.680 --> 06:34:09.200 the cube root of function, which means the cube root function 06:34:09.200 --> 06:34:16.680 cubed, which gets us back to x. Similarly, if we compute f inverse of f of x, that's 06:34:16.680 --> 06:34:19.610 the cube root of x cubed. 06:34:19.610 --> 06:34:25.600 And we get back to excellence again. So the cube root function really is the inverse of 06:34:25.599 --> 06:34:30.699 the cubing function. When we compose the two functions, we get back to the number that 06:34:30.700 --> 06:34:37.440 we started with. It'd be nice to have a more systematic way of finding inverses of functions 06:34:37.439 --> 06:34:44.547 besides guessing and checking. One method uses the fact that inverses reverse the roles 06:34:44.547 --> 06:34:50.520 of y and x. So if we want to find the inverse of the function, f of x equals five minus 06:34:50.520 --> 06:34:59.840 x over 3x. We can write it as y equals five minus x over 3x. Reverse the roles of y and 06:34:59.840 --> 06:35:09.957 x To get x equals five minus y over three y, and then solve for y. To solve for y, let's 06:35:09.957 --> 06:35:19.250 multiply both sides by three y. Bring all terms with y's in them to the left side, and 06:35:19.250 --> 06:35:28.369 alternate without wizened them to the right side, factor out the why. And divide to isolate 06:35:28.369 --> 06:35:40.047 why this gives us f inverse of x as five over 3x plus one. Notice that our original function 06:35:40.047 --> 06:35:46.307 f and our inverse function, f inverse are both rational functions, but they're not the 06:35:46.308 --> 06:35:53.080 reciprocals of each other. And in general, f inverse of x is not usually equal to one 06:35:53.080 --> 06:36:01.350 over f of x. This can be confusing, because when we write two to the minus one, that does 06:36:01.349 --> 06:36:08.750 mean one of our two, but f to the minus one of x means the inverse function and not the 06:36:08.750 --> 06:36:16.047 reciprocal. It's natural to ask us all functions have inverse functions. That is for any function 06:36:16.047 --> 06:36:23.119 you might encounter. Is there always a function that its is its inverse? In fact, the answer 06:36:23.119 --> 06:36:30.279 is no. See, if you can come up with an example of a function that does not have an inverse 06:36:30.279 --> 06:36:37.977 function. The word function here is key. Remember that a function is a relationship between 06:36:37.977 --> 06:36:47.559 x values and y values, such that for each x value in the domain, there's only one corresponding 06:36:47.560 --> 06:36:57.388 y value. One example of a function that does not have an inverse function is the function 06:36:57.387 --> 06:37:01.067 f of x equals x squared. 06:37:01.067 --> 06:37:02.237 To see that, 06:37:02.238 --> 06:37:09.840 the inverse of this function is not a function. Note that for the x squared function, the 06:37:09.840 --> 06:37:17.957 number two and the number negative two, both go to number four. So if I had an inverse, 06:37:17.957 --> 06:37:27.029 you would have to send four to both two and negative two, the inverse would not be a function, 06:37:27.029 --> 06:37:34.217 it might be easier to understand the problem, when you look at a graph of y equals x squared. 06:37:34.218 --> 06:37:40.878 Recall that inverse functions reverse the roles of y and x and flip the graph over the 06:37:40.878 --> 06:37:47.840 line y equals x. But when I flipped the green graph over the line y equals x, I get this 06:37:47.840 --> 06:37:53.599 red graph. This red graph is not the graph of a function because it violates the vertical 06:37:53.599 --> 06:37:59.859 line test. The reason that violates the vertical line test is because the original green function 06:37:59.860 --> 06:38:10.000 violates the horizontal line test, and has 2x values with the same y value. In general, 06:38:10.000 --> 06:38:14.950 a function f has an inverse function if and only if the graph of f satisfies the horizontal 06:38:14.950 --> 06:38:21.690 line test, ie every horizontal line intersects the graph. In it most one point, pause the 06:38:21.689 --> 06:38:26.869 video for a moment and see which of these four graphs satisfy the horizontal line test. 06:38:26.869 --> 06:38:34.579 In other words, which of the four corresponding functions would have an inverse function? 06:38:34.580 --> 06:38:42.160 You may have found that graphs A and B violate the horizontal line test. So their functions 06:38:42.159 --> 06:38:48.860 would not have inverse functions. But graph C and D satisfy the horizontal line test. 06:38:48.860 --> 06:38:54.740 So these graphs represent functions that do have inverses. functions that satisfy the 06:38:54.740 --> 06:39:02.399 horizontal line test are sometimes called One to One functions. Equivalently a function 06:39:02.399 --> 06:39:10.739 is one to one, if for any two different x values, x one and x two, the y value is f 06:39:10.739 --> 06:39:18.530 of x one and f of x two are different numbers. Sometimes, as I said, f is one to one, if, 06:39:18.529 --> 06:39:27.509 whenever f of x one is equal to f of x two, then x one has to equal x two. As our last 06:39:27.509 --> 06:39:32.939 example, let's try to find P inverse of x, where p of x is the square root of x minus 06:39:32.939 --> 06:39:40.707 two drawn here. If we graph P inverse on the same axis as p of x, we get the following 06:39:40.707 --> 06:39:49.457 graph simply by flipping over the line y equals x. If we try to solve the problem algebraically 06:39:49.457 --> 06:39:56.728 we can write y equal to a squared of x minus two, reverse the roles of y and x and solve 06:39:56.728 --> 06:40:07.579 for y by squaring both sides adding two. Now if we were to graph y equals x squared plus 06:40:07.580 --> 06:40:13.290 two, that would look like a parabola, it would look like the red graph we've already drawn 06:40:13.290 --> 06:40:22.670 together with another arm on the left side. But we know that our actual inverse function 06:40:22.669 --> 06:40:31.779 consists only of this right arm, we can specify this algebraically by making the restriction 06:40:31.779 --> 06:40:40.897 that x has to be bigger than or equal to zero. This corresponds to the fact that on the original 06:40:40.898 --> 06:40:47.530 graph for the square root of x, y was only greater than or equal to zero. Looking more 06:40:47.529 --> 06:40:55.820 closely at the domain and range of P and P inverse, we know that the domain of P is all 06:40:55.820 --> 06:41:01.650 values of x such that x minus two is greater than or equal to zero. Since we can't take 06:41:01.650 --> 06:41:07.250 the square root of a negative number. This corresponds to x values being greater than 06:41:07.250 --> 06:41:14.069 or equal to two, or an interval notation, the interval from two to infinity. The range 06:41:14.069 --> 06:41:20.419 of P, we can see from the graph is all y value is greater than or equal to zero, or the interval 06:41:20.419 --> 06:41:29.047 from zero to infinity. Similarly, based on the graph, we see the domain of P inverse 06:41:29.047 --> 06:41:34.137 is x values greater than or equal to zero, the interval from zero to infinity. And the 06:41:34.137 --> 06:41:40.500 range of P inverse is Y values greater than or equal to two, or the interval from two 06:41:40.500 --> 06:41:45.907 to infinity. If you look closely at these domains and ranges, you'll notice that the 06:41:45.907 --> 06:41:53.919 domain of P corresponds exactly to the range of P inverse, and the range of P corresponds 06:41:53.919 --> 06:41:57.569 to the domain of P inverse. 06:41:57.569 --> 06:42:04.387 This makes sense, because inverse functions reverse the roles of y and x. The domain of 06:42:04.387 --> 06:42:10.977 f inverse of x is the x values for F inverse, which corresponds to the y values or the range 06:42:10.977 --> 06:42:18.349 of F. The range of f inverse is the y values for F inverse, which correspond to the x values 06:42:18.349 --> 06:42:28.559 or the domain of f. In this video, we discussed five key properties of inverse functions. 06:42:28.560 --> 06:42:36.920 inverse functions, reverse the roles of y and x. The graph of y equals f inverse of 06:42:36.919 --> 06:42:47.349 x is the graph of y equals f of x reflected over the line y equals x. When we compose 06:42:47.349 --> 06:42:56.099 F with F inverse, we get the identity function y equals x. And similarly, when we compose 06:42:56.099 --> 06:43:04.217 f inverse with F, that brings x to x. In other words, F and F inverse undo each other. The 06:43:04.218 --> 06:43:18.218 function f of x has an inverse function if and only if the graph of y equals f of x satisfies 06:43:18.218 --> 06:43:31.110 the horizontal line test. And finally, the domain of f is the range of f inverse and 06:43:31.110 --> 06:43:39.000 the range of f is the domain of f inverse. These properties of inverse functions will 06:43:39.000 --> 06:43:45.457 be important when we study exponential functions and their inverses logarithmic functions.