If a function is everywhere continuous and differentiable such that $f^{\prime}(x) \geq 6$ for all $x \in[2,4]$ and $f(2)=-4$, then

$f(4) \geq 8$

Since $f(x)$ is everywhere continuous and differentiable. Therefore, by Lagrange's mean value theorem there exists $c \in(2,4)$ such that

$f^{\prime}(c)=\frac{f(4)-f(2)}{4-2}$

$\Rightarrow \frac{f(4)+4}{2} \geq 6$               $\left[f'(x) \geq 6\right.$ for all $\left.x \in[2,4]\right]$

$\Rightarrow f(4) \geq 8$