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How can I demonstrate that if \[ g\colon\mathbb R\to \mathbb R, x \mapsto f(x) + \frac 12\arctan{\sqrt{x+1}} \] is constant, where \[ f\colon\mathbb R\to\mathbb R, x\mapsto \arctan(x+2) - \arctan x. \]

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I TeXified your input, please check if I did correctly. –  martini Sep 19 '12 at 16:43
    
yes , it is :) , thank you –  Cioroianu Denis Sep 19 '12 at 16:43
    
the question is constant –  Cioroianu Denis Sep 19 '12 at 16:51
    
Simply derive, and find that $g'=0$. –  Lucien Sep 19 '12 at 17:01
    
Substitution $f$ in $g$ yields $$g(x)=\arctan(x+2) - \arctan x+ \frac 12\arctan{\sqrt{x+1}},$$ so for verify if $f$ is constant, you may differentiate it, because composition $g\circ f$ is differetiable (prove this fact). –  M. Strochyk Sep 19 '12 at 17:06
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1 Answer

It is not constant.

Differentiate to get $g'(x) = \frac{1}{(x+2)^2+1} - \frac{1}{x^2+1}+\frac{1}{4 \sqrt{x+1}(x+2)}$. $g'(0) = -\frac{27}{4}$, hence it is not constant.

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