#### Answer

$$ f(x) \geq10 x-15$$

#### 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 $[2,x]$, we get
\begin{align*}
\frac{f(x)-f(2)}{x-2}&=f^{\prime}(c)\\
f(x)-f(2)&=(x-2) f^{\prime}(c)\\
\end{align*}
Since $f'(x) \geq10$, then
\begin{align*}
f(x)-f(2) &\geq 10(x-2), \\
f(x)& \geq f(2)+10(x-2)\\
&\geq10 x-15
\end{align*}