Answer
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,4]$
On the interval $(1, 4)$, $f'(c) = \frac{ln(4)}{3}$ when $c = \frac{3}{ln(4)}$
Work Step by Step
$f(x) = ln~x$
1. The function is continuous on the interval $(0, \infty)$. Therefore, the function is continuous on the closed interval $[1,4]$
2. $f'(x) = \frac{1}{x}$ which exists for all numbers in the interval $(0, \infty)$. Therefore, $f'(x)$ exists for all numbers on the interval $(1, 4)$.
Thus $f$ is differentiable on the open interval $(1, 4)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[1,4]$
Therefore, according to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(1,4)$ such that $f'(c) = \frac{f(4)-f(1)}{4-1}$
$\frac{f(4)-f(1)}{4-1} = \frac{ln(4)-ln(1)}{3} = \frac{ln(4)}{3}$
We can find $c$:
$f'(x) = \frac{1}{x} = \frac{ln(4)}{3}$
$x=\frac{3}{ln(4)}$
On the interval $(1, 4)$, $f'(c) = \frac{ln(4)}{3}$ when $c = \frac{3}{ln(4)}$
