Calculus (3rd Edition)

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