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# 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]?

Mar 23, 2015

The conclusion of the Mean Value Theorem says that there is a number $c$ in the interval $\left(1 , 3\right)$ such that:
$f ' \left(c\right) = \frac{f \left(3\right) - f \left(1\right)}{3 - 1}$

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 $\left(1 , 3\right)$.

For $f \left(x\right) - - 2 {x}^{2} - x + 2$, we have
$f \left(1\right) = - 1$, and $f \left(3\right) = - 18 - 3 + 2 = - 19$
Also,
$f ' \left(x\right) = - 4 x - 1$.

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

$f ' \left(c\right) = - 4 c - 1 = \frac{f \left(3\right) - f \left(1\right)}{3 - 1} = \frac{- 19 - - 1}{3 - 1} = \frac{- 18}{2} = - 9$

So we need

$- 4 c - 1 = - 9$. 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..