How to prove that this function is bijective...

**gusztav** · April 13th 2009, 11:35 AM

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!

**Plato** · April 13th 2009, 11:57 AM

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

**gusztav** · April 13th 2009, 12:19 PM

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.

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