1 00:00:00,012 --> 00:00:03,969 [music]. The mean value theorem has some other 2 00:00:03,969 --> 00:00:11,342 applications where it connects derivative information back to information about the 3 00:00:11,342 --> 00:00:14,646 original function. Well, here's a theorem. 4 00:00:14,646 --> 00:00:19,332 Suppose the derivative of some function. So I'm assuming it's a differential 5 00:00:19,332 --> 00:00:23,562 function, is positive, it's greater than zero on some open interval. 6 00:00:23,562 --> 00:00:28,320 Then the function is an increasing function, meaning that it sends, bigger 7 00:00:28,320 --> 00:00:32,048 inputs to bigger outputs. You probably already believe that 8 00:00:32,048 --> 00:00:35,332 statement. But the mean value theorem let's us verify 9 00:00:35,332 --> 00:00:39,687 that it's actually true. Okay, so let's prove, this result. 10 00:00:39,687 --> 00:00:45,197 I'm going to pick, two points a and b, but I want to make sure that I pick them in 11 00:00:45,197 --> 00:00:48,243 order. So I'm going to pick b bigger than a so I 12 00:00:48,243 --> 00:00:51,131 can talk about the interval a b. All right. 13 00:00:51,131 --> 00:00:54,248 So I'll pick my two points. Now I'm going to apply the mean value 14 00:00:54,248 --> 00:00:58,057 theorem to the open interval a,b. And that's OK because I'm going to pick a 15 00:00:58,057 --> 00:01:02,152 and b to be in this open interval and that's mean that f being differentiable 16 00:01:02,152 --> 00:01:05,116 satisfies the conditions of the mean value theorem. 17 00:01:05,116 --> 00:01:09,812 The continuous on the closed interval a, b and it will be differnetial on the open 18 00:01:09,812 --> 00:01:10,951 interval a, b. Okay. 19 00:01:10,951 --> 00:01:17,383 So the mean value theorem then tells me to look at f of v minus f of a over b minus 20 00:01:17,383 --> 00:01:24,115 a, and the conclusion is that this is the derivative of f at some mystery point in 21 00:01:24,115 --> 00:01:27,892 between a and b. Now the good news is I don't know much 22 00:01:27,892 --> 00:01:31,656 about the derivative but I know the derivative is positive. 23 00:01:31,656 --> 00:01:35,280 So I know that f of b minus f of a over b minus a is positive. 24 00:01:35,280 --> 00:01:39,926 The other thing that I know is that b is greater than a and that means the 25 00:01:39,926 --> 00:01:42,927 denominator here b minus a is also positive. 26 00:01:42,927 --> 00:01:47,602 So I've got mystery number divided by positive is positive. 27 00:01:47,602 --> 00:01:51,435 That means this numerator must also be positive. 28 00:01:51,435 --> 00:01:56,974 So f of b minus f of a is positive. Now, if I add f of a to both sides, I get 29 00:01:56,974 --> 00:02:04,439 that f of b is bigger than f of a. If f sends bigger inputs to bigger outputs 30 00:02:04,439 --> 00:02:10,456 then f is increasing. Increasing means that a bigger input is 31 00:02:10,456 --> 00:02:15,438 sent to a bigger output. So if b is greater than a then f of b is 32 00:02:15,438 --> 00:02:21,710 bigger than f of a and this argument works for any a and b In the open interval on 33 00:02:21,710 --> 00:02:27,661 which I know the derivative is positive. So I'm going to conclude that f is an 34 00:02:27,661 --> 00:02:32,086 increasing function, which is exactly what I wanted to show. 35 00:02:32,086 --> 00:02:36,920 So if the derivative's positive on a whole interval, then the function's increasing 36 00:02:36,920 --> 00:02:40,130 on that interval. But it's important to point out that I 37 00:02:40,130 --> 00:02:44,551 needed to know the sign, the s i g n of the derivative on that entire interval. 38 00:02:44,551 --> 00:02:48,841 The mean value theorem ends of just picking out a single point, that point c, 39 00:02:48,841 --> 00:02:53,654 where it'll examine the derivative. But in order to cover all my bases right 40 00:02:53,654 --> 00:02:58,611 because I don't know what point c the mean value theorem might make me look at . 41 00:02:58,611 --> 00:03:03,636 I've got to control the sign the S I G N on the derivative on the entire interval. 42 00:03:03,636 --> 00:03:08,881 Alright, now I can play the same game when the sign of the derivative is negative. 43 00:03:08,881 --> 00:03:13,546 So, if I know the sign the SIG derivative is negative on some open interval then I 44 00:03:13,546 --> 00:03:18,288 include that F is a decreasing function. And, again this is just an application of 45 00:03:18,288 --> 00:03:20,884 the mean value theorem. Let's see the proof. 46 00:03:20,884 --> 00:03:24,754 So, I'm going to start out by again picking two points a,b in my opening 47 00:03:24,754 --> 00:03:29,170 interval and I'll again pick them so that b is greater than a so I can talk about 48 00:03:29,170 --> 00:03:32,702 the interval a,b. And then applying the mean value theorem, 49 00:03:32,702 --> 00:03:35,466 I should look at, f of b minus, f of a over b minus a. 50 00:03:35,466 --> 00:03:39,894 And the mean value theorem promises me that this is the derivative at some point, 51 00:03:39,894 --> 00:03:43,381 and I don't know what c is but it's the derivative at some point. 52 00:03:43,381 --> 00:03:48,061 But I know that the derivative at any point Is negative so this must be 53 00:03:48,061 --> 00:03:51,619 negative. Now b minus a is positive because b's 54 00:03:51,619 --> 00:03:55,852 bigger than a. So I've got a number I don't know divided 55 00:03:55,852 --> 00:04:01,340 by a positive number is negative. That means that f of b minus f of a must 56 00:04:01,340 --> 00:04:05,287 be negative. Then I add f of a to both sides and I can 57 00:04:05,287 --> 00:04:10,814 conclude that f of b is less than f of a. In other words, f sends bigger inputs to 58 00:04:10,814 --> 00:04:14,204 smaller output. Since f sends bigger inputs to smaller 59 00:04:14,204 --> 00:04:18,790 outputs, f is a decreasing function. Once again we're seeing that the mean 60 00:04:18,790 --> 00:04:23,681 value theorem lets me relate information about the derivative, in this case the 61 00:04:23,681 --> 00:04:26,748 fact that the sign of the derivative is negative. 62 00:04:26,748 --> 00:04:31,434 Back to information about the original function, in this case the function's 63 00:04:31,434 --> 00:04:34,746 decreasing. Alright, well lets summarize what we've 64 00:04:34,746 --> 00:04:38,010 learned thus far. So here are some applications of the mean 65 00:04:38,010 --> 00:04:40,822 value theorem. If f is differentiable on some open 66 00:04:40,822 --> 00:04:45,178 interval and depending as to what happens then we know something about f on that 67 00:04:45,178 --> 00:04:48,197 interval. So if the derivative is identically zero, 68 00:04:48,197 --> 00:04:52,418 if the derivative is equal to zero no matter what I plug in, then f is constant 69 00:04:52,418 --> 00:04:57,575 on that interval. If the derivative is positive no matter 70 00:04:57,575 --> 00:05:02,931 what I plug in, then f is increasing on that interval. 71 00:05:02,931 --> 00:05:10,455 And, if the derivative is negative, then f is decreasing on that interval.