The problem states: Justify if there exist a differentiable function f such that $|f(x)|>2$ and $f(x)f'(x)>sin(x)$ for every $x\in \mathbb{R}$. I thought about using trigonometric functions like $sin(x)+k$ or $arctan(x)$ but they do not fit on the second restriction $f(x)f'(x)>sin(x)$.

  • 1
    $\begingroup$ Do you recognise the expression $f(x) f'(x)$? It has a nice antiderivative, which may help you make some progress. $\endgroup$ Commented May 13 at 14:08
  • $\begingroup$ Hi Miguel, I've rolled back your edit. In general if your question had a reasonable interpretation and someone has put time into writing you a good answer, you shouldn't change your question in a way that makes their answer invalid - instead you should thank them, and post a new question with the change (and make sure you proofread it!) $\endgroup$ Commented May 14 at 11:17

1 Answer 1


Notice that $f(x)f'(x)=(\frac 12f(x)^2)'$

Also you can always reinterpret $a>b$ into $a=b+c$ with $c>0$. Here we are not dealing with number but functions therefore we want:

$(\frac 12f(x)^2)'=\sin(x)+c(x)\implies \frac 12f(x)^2=C(x)-\cos(x)$

Where $C(x)$ is an antiderivative of $c(x)>0$

  • Since you want $|f|>2$ i.e. $|\frac 12f^2|>2$ on whole $\mathbb R$ you need $C(x)>3$

$C'(x)=c(x)>0$ so $C$ strictly increasing on whole domain and positive, that's when exponentials get handy.

An easy way to fulfil this condition is therefore $c(x)=e^x$ and $C(x)=3+e^x$

Draw $f(x)=\sqrt{6+2e^x-2\cos(x)}$ and verify it solves the exercise


The case $|f|<2$ is impossible !

You would need $0<C(x)-\cos(x)<2\implies 1<C(x)<1$.

Relaxing the condition to let say $|f|<\sqrt{8}$ (this is just for simple coefficients afterwards, anything greater than $2$ would fit).

You end up with $1<C(x)<3\implies C(x)-2\in(-1,1)$ increasing and the function $\tanh(x)$ comes in mind.


Now $f(x)=\sqrt{4+2\tanh(x)-2\cos(x)}\ $ works

  • 1
    $\begingroup$ Hello. I'm not sure I understand why the constant has to be greater than $3$. It seems to me that it's fine if it's equal to $3$, because the strictness comes for free anyway from the fact that $e^x > 0$, right? $\endgroup$ Commented May 13 at 22:23
  • $\begingroup$ @zwim Sorry but made an edit, wanted to write $|f(x)|<2$. $\endgroup$
    – MiguelCG
    Commented May 14 at 10:13
  • 1
    $\begingroup$ @IzaakvanDongen You are totally right, I was too conservative and forgot $e^x$ could solve the strict inequality by itself. Edited accordingly. $\endgroup$
    – zwim
    Commented May 14 at 11:02
  • $\begingroup$ @zwim Thank you for all your work on the post, after all, I made a mistake on the problem statement but the conclusion you have shown to me are the root of what I was looking for. $\endgroup$
    – MiguelCG
    Commented May 14 at 13:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .