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 $$[1,9]$$.