Answer
$$f(4) \geq 8$$
Work Step by Step
By using the MVT, there exists $c\in[a,b]$ such that
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
$$
Applying on $[2,4]$
\begin{align*}
\frac{f(4)-f(2)}{4-2}&=f^{\prime}(c)\\
\frac{f(4)-(-2)}{2}& \geq 5\\
f(4)&\geq 8
\end{align*}