Proving inequality for derivative relationship

Question

I have a differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(0)=0$ and $|f'(x)|\le 1$ $\forall x\in\mathbb{R}$. Now I want to prove that: $$-2\le f(2)\le 2.$$ So from the given relationship for $x=2$ I get: $$-1\le\lim_{x\to2}{\frac{f(x)-f(2)}{x-2}}\le 1\iff-1\le\frac{f(x)-f(2)}{x-2}\le1$$ But how can I prove the thing that I want? I am thinking of using the squeeze theorem. Any hints?

Answer 1

If $f(2)>2$ then by the Mean Value Theorem there is $t\in (0,2)$ such that $$2<f(2)-f(0)=f'(t)(2-0)\implies f'(t)>1$$ which is a contradiction since $|f'(x)|\leq 1$ for all $x$. What about the other inequality $f(2)<-2$?

Answer 2

Use the mean value theorem:$$\bigl|f(2)\bigr|=2\left|\frac{f(2)-f(0)}{2-0}\right|=\bigl|f'(c)\bigr|,$$for some $c\in(0,2)$.