Verifying MVT hypotheses and finding the guaranteed value of c

Since $\sqrt{x}$ is continuous for all $x \ge 0$, $f$ is continuous on $[0,4]$. Its derivative $f'(x) = \frac{1}{2\sqrt{x}}$ exists for all $x>0$, so $f$ is differentiable on the open interval $(0,4)$ (the only point where $f'$ fails to exist is $x=0$, which is not in the open interval). Thus both hypotheses of the MVT are satisfied, so the theorem guarantees at least one $c$ in $(0,4)$ with $f'(c) = \frac{f(4)-f(0)}{4-0}$.

The secant slope is $\frac{f(4)-f(0)}{4-0} = \frac{2-0}{4} = \frac{1}{2}$. Setting $f'(c) = \frac{1}{2\sqrt{c}} = \frac{1}{2}$ gives $\sqrt{c} = 1$, so $c = 1$. Since $c=1$ lies in $(0,4)$, this is the value guaranteed by the Mean Value Theorem.

The tangent line at c is parallel to the secant line joining the endpoints. This is because the Mean Value Theorem guarantees a point where the instantaneous slope equals the average slope over the interval. Equal slopes mean the two lines never intersect, so they are parallel.