2
$\begingroup$

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 .

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

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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