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

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

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

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

is a bijective function?

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

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

Thanks!

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

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

Of course,

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

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

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

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