# How do you determine all values of c that satisfy the mean value theorem on the interval [-1,1] for (x^2 − 9)(x^2 + 1)?

Jan 25, 2018

$c = 0$

#### Explanation:

First, we need to figure out what the points are, so we know what the slope between them is.

$f \left(- 1\right) = \left({\left(- 1\right)}^{2} - 9\right) \left({\left(- 1\right)}^{2} + 1\right) = \left(1 - 9\right) \left(1 + 1\right) = - 16$

$f \left(1\right) = \left({\left(1\right)}^{2} - 9\right) \left({\left(1\right)}^{2} + 1\right) = \left(1 - 9\right) \left(1 + 1\right) = - 16$

The two points have the same value, so the slope between them is zero.

The mean value theorem says that:

If the slope between two points on a graph is $m$, then there must be some point $c$ between those points at which the derivative is also $m$.

In this case, our slope is 0, so we're looking for points between $- 1$ and $1$ whose slope(s) is/are 0.

To do this, let's take the derivative of our function.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({x}^{2} - 9\right) \left({x}^{2} + 1\right)$

$= 2 x \left({x}^{2} + 1\right) + 2 x \left({x}^{2} - 9\right)$

$= 2 {x}^{3} + 2 x + 2 {x}^{3} - 18 x$

$= 4 {x}^{3} - 16 x$

This derivative represents our slope, so we can set it equal to 0, since we're looking for where the slope is 0.

$4 {x}^{3} - 16 x = 0$

${x}^{3} - 4 x = 0$

$x \left({x}^{2} - 4\right) = 0$

$x \left(x - 2\right) \left(x + 2\right) = 0$

$\therefore x \in \left\{- 2 , 0 , 2\right\}$

The three points where the slope is zero are $- 2$, $0$, and $2$. However, since our problem wants us to find points we can use for the MVT for $- 1$ and $1$, we can only choose points between $- 1$ and $1$. Therefore, the only point we can use is 0.

$c = 0$