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## Differential Calculus (2017 edition)

### Unit 11: Lesson 7

Mean value theorem

# Mean value theorem example: polynomial

Sal finds the number that satisfies the Mean value theorem for f(x)=x²-6x+8 over the interval [2,5]. Created by Sal Khan.

## Video transcript

Let's say I have some function f of x that is defined as being equal to x squared minus 6x plus 8 for all x. And what I want to do is show that for this function we can definitely find a c in an interval where the derivative at the point c is equal to the average rate of change over that interval. So let's give ourselves an interval right over here. Let's say we care about the interval between 2 and 5. And this function is definitely continuous over this closed interval, and it's also differentiable over it. And it just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, that's in the open interval such that the derivative at c is equal to the average rate of change over this interval. Or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see f of 5 minus f of 2, f of 5 is, let's see, f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8. That's going to be a 0. So this is equal to 3/3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be in this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7/2, which is the same thing as 3 and 1/2. So it's definitely in this interval right over here. So we've just found our c is equal to 7/2. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis. And then this right over here is our x-axis. Looks like all the action is happening in the first and fourth quadrants. So that is our x-axis. And let's see, let's say this is 1, 2, 3, 4, and 5. So we already know that 2 is one of the zeroes here. So we know that our function if we wanted to graph it, it intersects the x-axis right over here. And you can factor this out as x minus 2 times x minus 4. So the other place where our function hits 0 is when x is equal to 4 right over here. Our vertex is going to be right in between at x is equal to 3. When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1. So you have the point 3, negative 1 on it. And so that's enough for us to graph. And we also know at 5, we're at 3. So 1, 2, 3. So at 5, we are right over here. So over the interval that we care about, our graph looks something like this. So that's the interval that we care about. And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points. And if I were to just visually look at it, I'd say well yeah, it looks like right around there just based on my drawing, the slope of the tangent line looks like it's parallel. It looks like it has the same slope, looks like the tangent line is parallel to the secant line. And that looks like it's right at 3 and 1/2 or 7/2. So it makes sense. So this right over here is our c. c is equal to 7/2.