How prove $|f'(x)|\le 4$ if $|f(x)|\le 1$ 
Let $f(x)$ be differentiable on $\Bbb R$, and for any $x_{0}\in \Bbb R$, 
  $$0<f'(x_{0}+x)-f'(x_{0})<4x \qquad(x>0)$$
  and if $|f(x)|\le 1$, show that
  $|f'(x)|\le 4$.

I tried to use Langrange’s theorem, but I couldn’t.
This is an exam problem from yesterday.
Thank you for you help!
 A: I suppose there is some issue in the question. If the given condition on $f'(x)$ holds for any $x_{0} \in \mathbb{R}$ then it is easy to obtain a contradiction.
From the given condition on $f'(x)$ it follows that $f'(x)$ is strictly increasing. Hence either $\lim_{x \to \infty}f'(x) = L\text{ or }\infty$ and $\lim_{x \to -\infty}f'(x) = M\text{ or }-\infty$. We will need to show that both these limit $L, M$ exist and are finite and both have absolute value less than equal to $4$.
First we need to see why $f'(x)$ can't tend to $\infty$ as $x \to \infty$. Clearly $f(x) - f(x/2) = (x/2)f'(c)$ for $0 < x/2 < c < x$ and we can see that as $x \to \infty$ LHS is bounded and RHS is not. Hence $f'(x) \to L$ as $x \to \infty$. Similarly $f'(x) \to M$ as $x \to -\infty$. From the same equation it is easy to see that $L = M = 0$ and hence $f'(x)$ is a constant and we get a contradiction.
I believe the question should be that there is some specific point $x_{0} \in \mathbb{R}$ for which we have $0 < f'(x_{0} + x) - f'(x_{0}) < 4x$ for all $x > 0$.
Update: If we assume that the condition on $f'(x)$ applies only to a specific point $x_{0}$ then it effectively constrains the values of $f'(x)$ for $x > x_{0}$. But at the same time it does not say anything about the values of $f'(x)$ for $x < x_{0}$. Typically it is easily to construct a bounded function $f(x)$ with $|f(x)| \leq 1$ but with its derivative being unbounded. A case in point is $\sin(1/x)$. So I believe it is possible to create a counterexample to $|f'(x)|\leq 4$ for values of $x < x_{0}$. Can you please check the source of the question you have asked and confirm that there is no mistake in the question.
