Inverse function of a piecewise polynomial

Question

What is the inverse function of $f:\mathbb R\to\mathbb R$ with $$f(x)=\begin{cases} x^2&x\ge0\\ -x&x<0\end{cases}$$

I have trouble finding inverse function. For some outputs such as $25$ I am not able to determine which condition did it satisfy.

Answer 1

The function is not bijective so it doesn't have an inverse.

Answer 2

Answering your question about: 'why does that mean it does not have an inverse'? If there are two or more $x-$ values for every $y$-value, then the inverse function will have two or more $y-$ values for every $x$-value. The inverse function does not pass the vertical line test so it is not considered a function.

An alternative is to restrict the domain so that the function has an inverse. We can do this with the functions on either side of $x = 0$, since the function has a unique $x$-value for every $y$-value.

Geometrically speaking, the inverse function is the reflection across the line $y=x$, which is equivalent to 'swapping' $x$ and $y$.

With the domain $x ≤ 0$, the inverse function is $x = -y$ or $y = -x$ where $y ≤ 0$.

With the domain $x ≥ 0$, the inverse function will be $x = y^2$ where $y ≥ 0$. Since $y ≥ 0$, we can also rewrite this as $y = \sqrt{x}$.

Answer 3

$f$ only assumes positive values, and for each $p > 0$ there are two points mapping to it: $\sqrt{p}>0$ and $-p < 0$.

So $f$ does not have an inverse as a map $\Bbb R \to \Bbb R$ or even $\Bbb R \to [0,+\infty)$, It does have an inverse when we restrict the domain to $(-\infty,0]$ and the codomain to $[0,+\infty)$, namely $f(x)=-x$ again, and also with domain and codomain both equal to $[0,+\infty)$, namely $f(x)=\sqrt{x}$.