Let $f(x)=\sqrt{x}$ on $[0,9]$. The Mean Value Theorem guarantees a number $c$ in $(0,9)$ with $f'(c)=\dfrac{1}{3}$. What is $c$?

Core Concept

$f'(x)=\frac{1}{2\sqrt{x}}$. Setting $\frac{1}{2\sqrt{c}}=\frac{1}{3}$ gives $\sqrt{c}=\frac{3}{2}$, so $c=\frac{9}{4}$.