Question

Let $f:R\to R$ be defined as $f\left(x\right)=x^5$, show that it is a bijective function.

Answer 1

$f\left(x\right)=x^5$

$y=f\left(x\right)=x^5$

$x=f^{-1}=\sqrt[5]{5y}$

Because $f^{-1}$ is defined for all $y\in R$, we have that f is surjective or onto.

The implication

$y_1=y_2\Rightarrow f^{-1}\left(y_1\right)=f^{-1}\left(y_2\right)$

entails that f is injective or one-one.

Being surjective and injective it is bijective (one-to-one).