# Given the function f(x)=-x^3+4x^2-3, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [0,4] and find the c?

Jan 26, 2017

The values of $c$ are $= \left\{0 , \frac{8}{3}\right\}$

#### Explanation:

$f \left(x\right)$ is a polynomial function.

So it is continuous on the interval $\left[0 , 4\right]$and differentiable on the interval ]0,4[, therefore we can apply the mean value theorem which states that

there is $c \in \left[0 , 4\right]$ such that

$f ' \left(c\right) = \frac{f \left(4\right) - f \left(0\right)}{4 - 0}$

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

$f \left(0\right) = - 0 + 0 - 3 = - 3$

$f \left(4\right) = - 64 + 64 - 3 = - 3$

Therefore,

$f ' \left(c\right) = \frac{f \left(4\right) - f \left(0\right)}{4 - 0} = \frac{- 3 - \left(- 3\right)}{4} = 0$

Also,

$f ' \left(x\right) = - 3 {x}^{2} + 8 x$

$f ' \left(c\right) = - 3 {c}^{2} + 8 c = 0$

Solving for $c$

$c \left(- 3 c + 8\right) = 0$

So,

$c = 0$

and

$c = \frac{8}{3}$

Both values of $c \in \left[0 , 4\right]$