Question

How do you give a value of c that satisfies the conclusion of the Mean Value Theorem for Derivatives for the function `f(x)=-2x^2-x+2` on the interval [1,3]?

Answer 1

The conclusion of the Mean Value Theorem says that there is a number `c` in the interval `(1, 3)` such that:
`f'(c)=(f(3)-f(1))/(3-1)`

To find (or try to find) `c`, set up this equation and solve for `c`.
If there's more than one `c` make sure you get the one (or more) in the interval `(1, 3)`.

For `f(x)--2x^2-x+2`, we have
`f(1)=-1`, and `f(3)=-18-3+2=-19`
Also,
`f'(x)=-4x-1`.

So the `c` we're looking for satisfies:

`f'(c)=-4c-1=(f(3)-f(1))/(3-1)=(-19--1)/(3-1)=(-18)/2=-9`

So we need

`-4c-1=-9`. And `c=2`.

Note:
I hope you've been told that actually finding the value of `c` is not a part of the Mean Value Theorem.
The additional question"find the value of `c`" is intended as a review of your ability to solve equations. For most functions, you will not be able to find the `c` that the MVT guarantees us is there..