# Thread: How to prove that this function is bijective...

1. ## How to prove that this function is bijective...

Hi, how can I prove that $f:\mathbb{R} \to \mathbb{R}$, where

$f(x)=x^3+4x-\cos(\pi x)$

is a bijective function?

If the function were just $g(x)=x^3+4x$, the problem could be easily solved by noting that $x^3$ and $x$ are strictly increasing functions, and their linear combination is therefore also strictly increasing. As the image of $g(x)$ is whole $\mathbb{R}$, $g(x)$ is bijective.

But in $f(x)$, we also have $\cos(\pi x)$, which is a periodic, and not strictly monotonous function. Do we have to show that $f(x)$ is simultaneously injective and surjective to be a bijection or is there some other way?

Thanks!

2. $f'(x) = 3x^2 + 4 + \pi \sin (\pi x)$.
What does that tell you?

3. Originally Posted by Plato
$f'(x) = 3x^2 + 4 + \pi \sin (\pi x)$.
What does that tell you?
Aha! Plato, thank you so much!

Of course,

$\sin (\pi x) \in [-1,1], \forall x \in \mathbb{R}$,

$\pi \sin (\pi x) \in [-\pi,\pi], \forall x \in \mathbb{R}$

and naturally $4 + \pi \sin (\pi x) > 0, \forall x \in \mathbb{R}$

And because $x^2$ is also positive, f $'(x)>0, \forall x \in \mathbb{R}$, so our $f$ must be a strictly increasing function.