Prove that the function $f\colon \mathbb{R} \to \mathbb{R}$ , $f(x) = x*|x|$ is a bijection

Question

I needed help proving this

This is what I tried, I broke it into two cases $x>0$ and $x<0$

Case 1 $x>0$

Injectivity

$|x|=x$

$f(x) = x^2$

$f(x)$ is a injection if $f(x)$ = $f(m)$ then $x=m$

$x^2 = m^2$

$x=m$ since $x>0$

Surjectivity

for every $x∈R$

$f(√x) = x $ therefore it is surjective

I needed help proving the case $x<0$

Answer 1

HINT: for $x\le 0$ you have that $f(x)=-x^2$.

Answer 2

For negative $x$ consider $f(\sqrt -x)$ and go from there.

Answer 3

It might be easier to view $$f(x) = \begin{cases} -x^2 &x < 0\\x^2&x \ge0\end{cases}.$$ So if z $<$ 0 , then z = -x$^2$ = -x (x) for some x $\in$ $\mathbb{R}$. And we proceed as you did.