I was practicing for a math test, and I got stuck at a calculus problem where I have two question that I couldn't solve for the past 3 days ..

The problem is as follow :

1st Problem

Given a function $F(x)$ where $x$ is a real number where and it's derivative : $$F'(x) = \frac{1}{x^2 + 1}$$

We don't have it's body but we know that $F(0) = 0$ and we do have : $$G(x) = F(x) + F(-x)$$ Find the derivative of G and compute G(0), prove that F(x) is odd .

My Take:

I computed G'(x) and found it's derivative :

$$G'(x) = \frac{2}{x^2 + 1}$$

I also computed $G(0) = 0$, but How can I prove that F(x) is odd in this case ? I tried to write this as following :

$F(-x) = G(x) - F(x)$ . I only know that it's odd for $x=0$ but how can I prove that it's odd for all real value that $x$ can take ?

2nd Problem:

My 2nd problem is that the practice problem tells me to prove that $H(x) = 2F(1)$ Given that $H(x) = F(x) + F(1/x)$

One interesting thing is that I"ve found is that it's derivative is always equal to 1 but I don't really have an idea on what I can do with this info .

Please clear up my confusion, I've been stuck on it for 3 days .

Thank's for your time gentlemen/ladies .

  • $\begingroup$ In problem 1, $F(x) = \arctan(x)$. $\endgroup$ – Austin Mohr Dec 5 '16 at 20:27

Answering the first point. By the chain rule, you rather have, for $x \in \mathbb{R}$, $$ G'(x) = F'(x) + (F(-x))'=F'(x)-F'(-x)=\frac{1}{x^2 + 1}-\frac{1}{(-x)^2 + 1}=0 $$ giving $G(x)=\text{constant}$ but from $G(0)=0$ one gets that $G(x)=0$, for all $x \in \mathbb{R}$, yielding $F(x) + F(-x)=0$ or $$F(-x)=-F(x), \quad x \in \mathbb{R} $$ meaning that $F$ is odd.

The second point. By the chain rule, you have, for $x \in \mathbb{R}$, $x \neq0$, $$ H'(x) = F'(x) + (F(1/x))'=F'(x)-\frac1{x^2}F'(1/x)= \cdots $$ Can you take from here?

  • $\begingroup$ Oh thank you so much, so my derivative was wrong that's why I couldn't solve it, it makes much more sense now, thank's :) . $\endgroup$ – Anis Souames Dec 5 '16 at 20:33
  • $\begingroup$ @AnisSouames You are welcome. $\endgroup$ – Olivier Oloa Dec 5 '16 at 20:34

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