Mean Value Theorem Application

For $f(x)=\sqrt{x}$, the average rate of change is 0.25, which means that $f'(x)$ is constant and equal to 0.25 for all $x$ in $[1,9]$.

Since $f(1)=1$ and $f(9)=3$, the average rate of change is $\frac{3-1}{9-1}=0.25$, so the instantaneous rate is 0.25 at both $x=1$ and $x=9$.

The Mean Value Theorem guarantees that there exists a $c=4$ in $(1,9)$ such that $f'(4)=0.25$, which is equal to the average rate of change of $f(x)$ on the interval.

The Mean Value Theorem does not apply because $f(x)=\sqrt{x}$ is not differentiable on $[1,9]$.