Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 \[ f\colon\mathbb R\to\mathbb R, x\mapsto \arctan(x+2) - \arctan x. \]

share|improve this question
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

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.