Differentiability: Mean Value Theorem

Since $f(x) = \sqrt{x-1}$ , $f'(x) = \frac{1}{2\sqrt{x-1}}$ . Let $c$ be the value of interest, then by the Mean Value Theorem there exists $c$ in the interval $[1,\ 38]$ that satisfies $\frac{f(38)-f(1)}{38-1} = \frac{1}{2\sqrt{c-1}}.$ Thus, $\frac{\sqrt{38-1}-\sqrt{1-1}}{38-1} = \frac{1}{2\sqrt{c-1}}$ $\; \Rightarrow \; \frac{\sqrt{37}}{37} = \frac{1}{2\sqrt{c-1}}$ $\; \Rightarrow \; c = \frac{41}{4}$ .

Since $c = \frac{41}{4}$ lies within the interval $[1,\ 38]$ , it satisfies the Mean Value Theorem.