Answer
On the graph, we can see that the tangent line at the point $(1,1)$ is parallel to the secant line.
Work Step by Step
$f(x) = \sqrt{x}$
$f'(x) = \frac{1}{2\sqrt{x}}$
1. $f$ is continuous on the interval $[0,4]$
2. $f$ is differentiable on the interval $(0,4)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,4]$
According to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(0,4)$ such that $f'(c) = \frac{f(4)-f(0)}{4-0}$
Then:
$f'(c) = \frac{f(4)-f(0)}{4-0}$
$\frac{1}{2\sqrt{c}} = \frac{\sqrt{4}-\sqrt{0}}{4-0}$
$\frac{1}{2\sqrt{c}} = \frac{2}{4}$
$2\sqrt{c} = 2$
$\sqrt{c} = 1$
$c = 1$
When $c = 1$, then $f(c) = \sqrt{1} = 1$
$(c,f(c)) = (1,1)$
On the graph, we can see that the tangent line at the point $(1,1)$ is parallel to the secant line.