Is there an inverse function of $f(x) = x^2 + \pi\cos x$?

Question

I don't think there is because of the $x^2$, but I don't know how to prove it.

Why is this getting so many downvotes? There's clearly an attempt to find a solution to the problem! — Shaun, Mar 12 '18 at 00:18
Note that as the question is posed the correct annswer should be “not sufficient information” or “it depends upon the domain” indeed $f:R\to R$ is not invertible but $f:[\pi/2,+\infty)\to[\pi^4/4,+\infty)$ is bijective and invertible. — user, Mar 12 '18 at 07:27
@Plzhelpme Please remember that you can choose an answer among the given if the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Mar 17 '18 at 22:44

score 3 · Answer 1 · edited Mar 12 '18 at 00:20

3

$$f (-x)=(-x )^2-\pi\cos (-x)= f (x)$$

$f $ is not injective and therefore it is not bijective.

So it does not have an inverse function.

edited Mar 12 '18 at 00:20

The Phenotype

5,199

answered Mar 12 '18 at 00:18

hamam_Abdallah

62,951

Thnks for editing. tnks a lot . – hamam_Abdallah Mar 12 '18 at 00:24

score 2 · Answer 2 · answered Mar 12 '18 at 00:40

2

It should be declared over which interval we are considering the function.

Indeed f(x) is not invertible as function from $\mathbb{R}\to\mathbb{R}$ but it is invertible, for example, if we assume a restriction $[a,+\infty)\to[b, +\infty)$ indeed

$f’(x)=2x-\pi\sin x>0$ for some $x\ge a$ and $f(a)=b$

and thus $f:[a,+\infty)\to[b, +\infty)$ is bijective.

answered Mar 12 '18 at 00:40

user

154,566

1

In fact $f'(x) > 0$ for $x > \pi/2$, so it's invertible from $[\pi/2, \infty)$ to $[\pi^2/4, \infty)$. – Robert Israel Mar 12 '18 at 03:00
@RibertIsrael Yes exactly since $\sin x<x$. – user Mar 12 '18 at 07:22

Is there an inverse function of $f(x) = x^2 + \pi\cos x$?

2 Answers2