# Consider the function f(x) = 4 x^3 − 4 x on the interval [ −3 , 3 ], how do you find the average or mean slope of the function on this interval?

Oct 26, 2017

See the explanation below

#### Explanation:

The average or mean slope of a function $f \left(x\right)$ on the interval $\left[a , b\right]$ is

$= \frac{f \left(b\right) - f \left(a\right)}{b - a}$

Here,

$f \left(x\right) = 4 {x}^{3} - 4 x = 4 x \left({x}^{2} - 1\right)$

and the interval is $= \left[- 3 , 3\right]$

$f \left(3\right) = 12 \left(8\right) = 96$

$f \left(- 3\right) = - 12 \cdot 8 = - 96$

Therefore,

The mean slope is

$= \frac{f \left(3\right) - f \left(3\right)}{3 - \left(- 3\right)} = \frac{96 + 96}{6} = 32$

Then,

There exists $c \in \left(- 3 , 3\right)$ such that

$f ' \left(c\right) = 32$

$f ' \left(x\right) = 12 {x}^{2} - 4 = 4 \left(3 {x}^{2} - 1\right)$

Therefore,

$f ' \left(c\right) = 4 \left(3 {c}^{2} - 1\right) = 32$

$3 {c}^{2} - 1 = 8$

${c}^{2} = \frac{9}{3} = 3$

$c = \pm \sqrt{3}$

So,

$\pm \sqrt{3} \in \left(- 3 , 3\right)$