If f([-1,2])=[2,8] and f is differentiable, prove there exists c$\in$[-1,2] such that |f'(c)| $\geq$2

Question

If f([-1,2])=[2,8] and f is differentiable, prove there exists c$\in$[-1,2] such that |f'(c)| $\geq$2

All I can see is that |f(2)-f(-1)| $\leq$ 6 and |2-(-1)| $\leq$ 3, so any $c_1$ and $c_2$ in [-1,2] will fulfill these 2 conditions, and that I probably need to use Mean Value Theorem.

Any help is apreciated, thanks!

Answer 1

• Hint 1: The image of $f$ is $[2,8]$, so there must be numbers $x,y\in[-1,2]$ such that $f(x)=2$ and $f(y)=8$.
• Hint 2: Indeed you need to use the mean value theorem. It states that there is an $a$ such that $f'(a)=\frac{f(x)-f(y)}{x-y}$. Can you argue why the magnitude of this fraction is $\geq2$?
• Remark: Note that both $x<y$ and $x>y$ are possible.