Show that $f(x) = \frac{x^3}{1+x^2}$ is bijective This seems like a simple question, but I'm stuck: how do I show that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x) = \frac{x^3}{1+x^2}$ is bijective?
I want to demonstrate that it is both injective and surjective. To show that it's injective, I need to show that $f(x) = f(y)$ implies $x = y$. However, I can't see a way to reduce $\frac{x^3}{1+x^2} = \frac{y^3}{1+y^2}$ to $x = y$ (since there are no like terms to combine). I'm also unsure of how to prove surjectiveness.

you don't need calculus to show that $f(x) = \dfrac{x^3}{1+x^2}$ is 1-1.
suppose $$\dfrac{a^3}{1+a^2} = \dfrac{b^3}{1+b^2} \tag 1$$ we will show that this implies $a = b$ proving $f$ is 1-1.
 cleaning up $(1)$ gives $$(a-b)(a^2 + ab + b^2 + a^2b^2) = 0$$ now use the fact that $a^2 + ab + b^2 > 0$ for $a \neq 0, b \neq 0$ to conclude $a = b.$
$\bf Edit:$ To show that $f$ is onto note that $f(x) = \dfrac{x^3}{1+x^2}$ is odd and $\lim_{x \to \infty} \dfrac{x^3}{1+x^2} = \infty$. That is, the range of $f$ is $(-\infty, \infty).$
