 In lecture 32 of our lecture series, we're talking about the mean value theorem, and I said when we first talked about the mean value theorem that the mean value theorem is probably the most important theorem from differential calculus. What makes the mean value theorem so important? Well, it turns out that the mean value theorem says in a nutshell that if you know the function's derivative, you basically know the function. That is, if we know the function's rate of change, we essentially know the function. We don't know all of the function. There is some important information, but we can infer a lot about a function if we know something. Show you those type of inferences we can make in this video. Suppose we have a function f so that its y-intercept is negative 3. And suppose we know that the function's derivative never gets bigger than 5 because it's less than or equal to 5 for all values of x. This would be saying something like I know the initial position of a moving particle and I know its velocity never gets sufficiently large. Could I make a guess where that particle is going to be the moment where time is 2? How large can f of 2 possibly be if under these assumptions here? Let's go to the mean value theorem. First notice that by assumption f is differentiable. So f prime of x is less than or equal to 5 for all values of x. That says it's differentiable, which if it's differentiable, that also implies it's continuous for all of these numbers. Those are the assumptions we need for the mean value theorem. Therefore the mean value theorem guarantees the following. It guarantees that there's some number c such that f prime of c is equal to the average rate of change here. That is, we're going to take f of 2 minus f of 0 over 2 minus 0, right? So what's going on here, f of 2, f of 0? Well, f of 2 is the number we're trying to predict and f of 0 is the last reference we actually know. That's the last instance of the function we have any data about. So can we make a prediction what's going to happen two units later? We'll let you know what it is. So we'll leave it alone for the moment. But f of 0 we do know that's going to be a negative 3. 2 minus 0, of course, is 2. And so we end up with this statement here that f of 2 plus 3 over 2, this is equal to f prime at c. But we do know something about the derivative. The derivative is less than or equal to 5. So this derivative f prime of c is less than or equal to 5. We know that. And so now we can start solving this inequality here for f of 2 times both sides of the inequality by 2. We get 10 is greater than or equal to f of 2 plus 3. So track 3 from both sides. We're going to get that 7 is greater than or equal to f of 2. Or if you prefer, we could write that as f of 2 is less than or equal to 7. And so that's an observation we can make. And in a retro spec, that might not be so surprising. If we think about it in terms of motion, right? If at time equals 0, the car is located at marker negative 3, right? And it's moving at most 5 units per time, 5 units of distance per time. And after 2 units of time, the most it could have traveled would be 10 units, 2 times 5. And if you take 10 plus negative 3, you get 7. So after 2 units of time, the farthest it could have traveled was 10 units. And therefore its current location is negative 3, we get 7, right? So that's usually my theorem, but it's kind of intuitive. It's like, okay, I get it. If we know a bound on its rate of change, that restricts how big the function could be. f of 2 could be at most 7, given this information. Let's look at another situation which we can make some inferences using the mean value theorem. Suppose that the derivative of a function satisfies the condition where the derivative can never get smaller than 2, but never bigger than 4 for all values x, okay? Well, let's find the best bounds for the numbers a and b. But we're trying to bound the quantity f of 10 minus f of 5. So how close together can f of 10 and f of 5 be? That gives us a. How far apart can they be? That gives us b. So how can we bound these numbers here? So what we're going to do is we will determine a and b using the mean value theorem. And we're going to consider the average rate of change going from 5 to 10. That is take f of 10 minus f of 5 over 10 minus 5. So that's what we're really interested in here. Because the mean value theorem would apply, we have a continuous differential function. This is equal to the derivative somewhere where a derivative is going to be bounded by 4 and bounded by 2. 10 minus 5, of course, is itself 5. So if we times both sides by 5, we're going to end up with 2 times 5, which is 10, which is less than or equal to f of 10 minus f of 5, which is less than or equal to 5 times 4, which is 20. And so given this information about the derivative, we can make the following inference. f of 10 minus f of 10 and 20. So we could take a to be the number 10. That's the smallest we can expect the difference to be. And then the biggest we can get is b. So what is f of 10 minus f of 5? So this is a net change. How did the function change from x equals 5 to x equals 10? Because we know the derivative fluctuates between 2 and 4. That is the derivative never gets outside of 2 and 4. This then tells us that the net change from f of 5 to f of 10, worst case scenario, it's 10, best case scenario, it's 20. It's bounded because the derivative is bounded. And these inferences we're making about our function f based upon its derivative is starting to suggest how much the derivative affects the shape of the original function f. And this is a consequence of the.