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

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 $x$ : $x^5 = y$.
