school exersise/ Differentiation

Question

If $f$ is differentiable in ${\bf R}$ and for every $x \in {\bf R}$, $$ f(x+\cos x)-f(1-x) \leq x\cos x , $$ then prove that $f'(1)=1/2$.

How is a school kid supposed to solve this exercise ? Thank you very much !

Answer 1

$$\lim_{x\to 0}\frac{f(x+\cos x)-f(1-x)}{x}=\lim_{x\to 0}{\bigg[(1-\sin x)f'(x+\cos x)+f'(1-x)\bigg]}=2f'(1)$$ We also have, for all $x>0\in\mathbb{R}$ that $$\frac{f(x+\cos x)-f(1-x)}{x}\le \cos x$$ and for all $x<0\in\mathbb{R}$ that $$\frac{f(x+\cos x)-f(1-x)}{x}\ge \cos x$$ These imply that $$\lim_{x\to 0^+}\frac{f(x+\cos x)-f(1-x)}{x}\le 1$$ and $$\lim_{x\to 0^-}\frac{f(x+\cos x)-f(1-x)}{x}\ge 1$$ respectively. But the limit exists, so these one-sided limits must agree and therefore $$2f'(1)=1$$ from which the claim follows.

Answer 2

By using the definition of the derivative and L'Hopital's rule to get

$$\begin{align} f'(1)=\lim_{x\to0}{f(1)-f(1-x)\over x}&\le\ge\lim_{x\to0}{f(1)+x\cos x-f(x+\cos x)\over x}\\ &=1+\lim_{x\to0}{f(1)-f(x+\cos x)\over x}\\ &=1-\lim_{x\to0}(1-\sin x)f'(x+\cos x)\\ &=1-f'(1) \end{align}$$

where, in the first line, the $\le$ sign applies when $x\gt0$ and the $\ge$ sign applies when $x\lt0$ (i.e., by taking the limit from the right for one inequality and from the left for the other).