Main content

## AP®︎/College Calculus AB

- Mean value theorem
- Mean value theorem example: polynomial
- Mean value theorem example: square root function
- Using the mean value theorem
- Justification with the mean value theorem: table
- Justification with the mean value theorem: equation
- Establishing differentiability for MVT
- Justification with the mean value theorem
- Mean value theorem application
- Mean value theorem review

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.

## Want to join the conversation?

- How did he get the vertex at 3?(8 votes)
- It appears he was relying on the symmetry of a parabola, as he says it’s halfway between the x-intercepts. There are other ways to get this however. Note that f'(x) = 2x – 6. Setting that equal to zero tells us the critical point is at x = 3, and in the case of a parabola, there is only one critical point, at the vertex.(33 votes)

- Would the average rate of change over an interval be the same as the value guaranteed by the Mean Value Theorem over the same interval?(2 votes)
- The Mean Value Theorem doesn’t guarantee any particular value or set of values. Rather, it states that for any closed interval over which a function is continuous, there exists some x within that interval at which the slope of the tangent equals the slope of the secant line defined by the interval endpoints.(10 votes)

- how does he find that f'(x) = 2x-6(1 vote)
`f(x) = x² - 6x +8`

`f'(x) = 2x - 6`

If you’ve watched the previous videos on taking derivatives, you should know the power rule which states:

if`f(x) = x^n`

, then`f'(x) = n*x^(n-1)`

.

You should also know the rule if`f(x) = g(x) + h(x)`

then`f'(x) = g'(x) + h'(x)`

(12 votes)

- At, why does Sal convert from brackets to parenthesis? Don’t those represent closed and open intervals? 0:55(1 vote)
- That’s correct. Recall that the statement of the mean value theorem requires that the function be continuous on the closed interval [a, b], but differentiable only on the open interval (a, b). Sal switches to open intervals when he begins talking about the derivative of the function.(5 votes)

- Why did he include the vertex on the graph about the mean theorem? plz explain this(1 vote)
- Sal did it to help himself graph the equation better.(5 votes)

- So does c represent the instant slope/rate of change/derivative? Or is it an x value were the derivative and secant slope are equal?(2 votes)
- c just represents the point where the instantaneous slope is equal to the average slope of the function in the closed interval.(3 votes)

- I’m just curious, if the equation wasn’t able to be factored, would two be the only value then that has a y-value of zero? I just want to make sure for future understanding. Thank you!(2 votes)
- If the equation is impossible/difficult to factor,

we would use the quadratic formula

to get the roots/zeros for the equation (read when y=0.. the x-intercept)(3 votes)

- If the equation is impossible/difficult to factor,
- How to calculate average f’ over a period (a,b)? Is this equal to average rate of change over (a,b)? Thanks.(2 votes)
- How is the x-value the same as a c?(1 vote)
- In the function
`f'(x) = 2x - 6`

,`x`

can be any value, and it would output a specific derivative of`f(x)`

.`c`

is the specific value of`x`

at which`f'(x)`

would be`1`

. In other words,`x`

can be any number that would give you a certain`f'(x)`

, but`c`

is the particular value that`x`

can take on to make`f'(x)`

be`1`

.(3 votes)

- In the function
- I was curious if the Number of points on the closed interval that satisfy the mean value theorem is = to the (number of inflection points) + 1(2 votes)
- Yes, you are correct!(1 vote)

## 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.