# Justify if there exist a differentiable function f such that $|f(x)|>2$ and $f(x)f'(x)>sin(x)$

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 $$f(x)f'(x)>sin(x)$$.

• Do you recognise the expression $f(x) f'(x)$? It has a nice antiderivative, which may help you make some progress. Commented May 13 at 14:08
• 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!) Commented May 14 at 11:17

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

https://www.desmos.com/calculator/2efsmgnqij

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

You would need $$0.

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 $$1 increasing and the function $$\tanh(x)$$ comes in mind.

$$\begin{cases}C(x)=2+\tanh(x)\\c(x)=1-\tanh(x)^2>0\end{cases}$$

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

• 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? Commented May 13 at 22:23
• @zwim Sorry but made an edit, wanted to write $|f(x)|<2$. Commented May 14 at 10:13
• @IzaakvanDongen You are totally right, I was too conservative and forgot $e^x$ could solve the strict inequality by itself. Edited accordingly.
– zwim
Commented May 14 at 11:02
• @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. Commented May 14 at 13:53