The table gives selected values of the differentiable function ${h}$.

Below is Tom’s attempt to write a formal justification for the fact that there exists a value ${c}$ in the interval $(-4,-3)$ such that $h^{\prime}({c})=2$.

Is Tom’s justification complete? If not, why?

Tom’s justification:

We are given that $h$ is differentiable, which means it’s both differentiable and continuous over the interval $[-4,-3]$. Furthermore, $\frac{h(-3)-h(-4)}{-3-(-4)}=2$

So, according to the mean value theorem, there exists a value ${c}$ somewhere in the interval $(-4,-3)$ such that $h^{\prime}(c)=2$. Options