# 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]?

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..