As $f(1)=-2 \& f^{\prime}(x) \geq 2 \forall x \in[1,6]$ Applying Lagrange's mean value theorem
$$
\begin{aligned}
&\frac{f(6)-f(1)}{5}=f^{\prime}(c) \geq 2 \\
&\Rightarrow f(6) \geq 10+f(1) \\
&\Rightarrow f(6) \geq 10-2 \\
&\Rightarrow f(6) \geq 8
\end{aligned}
$$