# Given the function f(x)=x(x^2-x-2), how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-1,1] and find the c?

Feb 13, 2018

$c = - \frac{1}{3}$

#### Explanation:

first of all, the Mean Value theorem isn't an hypothesis, it is a theorem which states that if a function is continuous over an interval $\left[a , b\right]$ and Differentiable on the interval $\left(a , b\right)$ then there exists a value c such that the slope of the tangent line at c is equal to the overall slope from a to b and $a < c < b$

and since this function is continuous over all values
, we know that there does exist such a value c, all we have to do is find it

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

where $a = - 1 \mathmr{and} b = 1$

therefore
$f ' \left(c\right) = - 1$

the derivative of this function can be found fairly simply by just using the power rule and the derivative is
$f ' \left(x\right) = 3 {x}^{2} - 2 x - 2$

therefore, to find c,
$3 {c}^{2} - 2 c - 2 = - 1$
$3 {c}^{2} - 2 c - 1 = 0$

and on solving this equation using the quadratic formula
we Get $c = - \frac{1}{3} \mathmr{and} 1$

but since c should be such that $a < c < b$, that is, c should be in the middle of the endpoints, $c = 1$ cannot be an answer as 1 is one of the endpoint.

Therefore
$c = - \frac{1}{3}$