# Given the function f(x)=(x-4)^2-1, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [3,0] and find the c?

Aug 14, 2016

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

#### Explanation:

Mean value theorem states that for a function defined and continuous on $\left[a , b\right]$ and continuously differentiable on $\left(a , b\right)$ then there exists some $a < c < b$ such that

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

With $a = 0 \mathmr{and} b = 3$ we have

$\frac{\left({\left(3 - 4\right)}^{2} - 1\right) - \left({\left(0 - 4\right)}^{2} - 1\right)}{3} = - 5$

$f ' \left(x\right) = 2 \left(x - 4\right)$

$\implies f ' \left(c\right) = 2 \left(c - 4\right)$

$2 c - 8 = - 5$

$2 c = 3 \implies c = \frac{3}{2}$