How to show the surjectivity of $f(x)=x^5$ on $\mathbb R$?

Question

Sasy $f:\mathbb R\to\mathbb R$ define by $f(x)=x^5$

This is definitely injective as $x_1^5=x_2^5 \implies x_1=x_2$

I say it is surjective because for all really $x$ there is all real $y$,

$x \in \mathbb R$ there $y \in \mathbb R$?

Answer 1

$f$ is continuous and $$\bullet\lim_{x\to-\infty }f(x)=-\infty$$ $$\bullet \lim_{x\to\infty }f(x)=+\infty.$$ By the intermediate value theorem, $f(\mathbb R)=\mathbb R$, then $f$ is surjective.

An other way to show the injectivity, is to remark that $$f'(x)=5x^4>0$$ for all $x\in\mathbb R\backslash \{0\}$, then the function is strickly increasing on $\mathbb R\backslash \{0\}$. By continuity on $\mathbb R$, we can conclude that $f$ is strinckly increasing on $\mathbb R$.

Answer 2

To show surjectivity, you show that for each $y$, there is an $x$ such that $f(x) = y$. This means you need to find an $x$ : $x^5 = y$.