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# Mean value theorem example: polynomial

AP.CALC:
FUN‑1 (EU)
,
FUN‑1.B (LO)
,
FUN‑1.B.1 (EK)

## 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 it for this function we can definitely find AC in an interval where the derivative at the point C is equal to the average rate of change over that interval so let's let's give ourselves an interval right over here let's say we care about the interval between 2 & 5 and this function is clearly both it's can definitely continuous over this closed interval and it's also it's also differentiable over and it just has to be differentiable over the open interval but this is differentiable really for all X and so let's show let's show that we can find 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 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'd encourage you to pause the video now and try to find AC 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 let me 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 0 so this is equal to 3 over 3 which is equal to 1 F prime of C is equal to 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 that's going to be neat we need to figure out what x value especially it has to be within this 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 and halves 7 halves 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 C is equal to 7 halves and let's just graph this to really make sure that this makes sense so let's let's 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 let's see let's say this is 1 2 3 4 & 5 so we already know that 2 is one of the zeros 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 our other 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 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 to graph it and we also know that 5 where 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 our graph looks something like this so that's the interval that we care about and we're saying that we were looking for 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 wow 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 three and a half or seven halves so it makes sense so this right over here is our CC is equal to seven halves
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