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$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

Thanks in advance :)

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    $\begingroup$ 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 $$ \frac{d^2}{dx^2}\arctan\log x = -\frac{(1+\log x)^2}{x^2(1+\log^2 x)^2}\leq 0.$$ $\endgroup$ – Jack D'Aurizio Sep 24 '18 at 20:25
  • $\begingroup$ That's not what the derivative of your function f is. $\endgroup$ – really Nobody Sep 24 '18 at 20:25
  • $\begingroup$ interesting question but your $f'(c)=1/(1+c^2)$ is not correct. $\endgroup$ – hamam_Abdallah Sep 24 '18 at 20:33
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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...

$$f'(x)\le f'(c)=\frac{f(x)-f(y)}{x-y} \le f'(y)$$

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  • $\begingroup$ i need more about it $\endgroup$ – KEVIN DLL Sep 24 '18 at 20:44
  • $\begingroup$ @KEVINDLL $x<c<y$ and $f'$ increasing thus... $\endgroup$ – hamam_Abdallah Sep 24 '18 at 21:19

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