# how to prove $f'(x) \leq \frac{f(x) - f(y)}{x - y} \leq f'(y)$

$$f(x) = x - \arctan(\ln(x))$$ on the interval $$[0,+\infty[$$. How to use the Mean value theorem to show that $$f'(x) \leq \frac{f(x) - f(y)}{x - y} \leq f'(y)$$, I know that according to theorem ,
$$\exists c \in ]x,y[, f'(c) = \frac{f(y) - f(x)}{y-x}$$
$$\frac{1}{1+c^2}= \frac{f(y) - f(x)}{y -x}$$
All help is appreciated

• I guess we are tacitly assuming $0<x<y$. And you just have to prove that $\arctan\log x$ is a concave function on $\mathbb{R}^+$, for instance by noticing that $$\frac{d^2}{dx^2}\arctan\log x = -\frac{(1+\log x)^2}{x^2(1+\log^2 x)^2}\leq 0.$$ – Jack D'Aurizio Sep 24 '18 at 20:25
• That's not what the derivative of your function f is. – really Nobody Sep 24 '18 at 20:25
• interesting question but your $f'(c)=1/(1+c^2)$ is not correct. – hamam_Abdallah Sep 24 '18 at 20:33

hint Here is the derivative of $f$ :
$$f'(x)=1-\frac{1}{x(1+\ln^2(x))}.$$ Observe that $$x\mapsto x(1+\ln^2(x))$$ is increasing at $(0,+\infty)$ as a product of two increasing functions. hence,
$f'$ is increasing thus...
$$f'(x)\le f'(c)=\frac{f(x)-f(y)}{x-y} \le f'(y)$$
• @KEVINDLL $x<c<y$ and $f'$ increasing thus... – hamam_Abdallah Sep 24 '18 at 21:19