Showing Surjectivitity of $f(x) = x^3$ I want to show that the function $f: \mathbb{R} \to \mathbb{R},\; f(x) = x^3$ is surjective.

First Question: If a function has an inverse, it is bijective yes?
Second Question: Is my process below, for showing surjectivity of $y = x^3$, incorrect because I am assuming that the function has an inverse? i.e., by assuming it has an inverse I am assuming it is bijective and using that to show it is surjective.

Surjective statement: For a function $f: A \to B$, the function is surjective if $\forall_{ y \in B}, \; \exists_{x \in A}\;|\;f(x) = y$
So then $\forall_{y \in \mathbb{R}}, \; \exists_{x \in \mathbb{R}}\; | \; f(x) = y $
If such an real $x$ exists, then we have $x^3 = y$ and $x = y^{\frac{1}{3}}$. $x$ is a real number (how do I show this?). It follows that $f(x) = (y^{\frac{1}{3}})^3 = y$

check f'(x) and the function will be strictly increasing and unbounded. so it will be surjective
