Find $f^{-1}(-3)$ if $f(x) =x|x|+1$

Question

So I can't somehow get the right answer and I was hoping that someone could correct me.

I was thinking that since the function involves absolute values I will have a piecewise function with two different conditions for x.

First I would like to state that $f(x)=y$ and $f^{-1}=y$

I could then proceed to employing the standard technique for inversing a function.

e.g $y=x|x|+1$

after switching x and y we get:

$x=y|y|+1$

because $|y|$ got two conditions

$|y|=-y$ if $y<0$

and

$|y|=y$ if $y\ge0$

we get two functions

(1) $x=y^2+1$ in terms of x we get $y=\sqrt {x-1}$ if $y\ge0$

and

(2) $x=-y^2+1$ in terms x we get $y=\sqrt {1-x}$ if $y<0$

I want the function when $f^{-1}(-3)$. Function (1) is underfined for $x=-3$ and function (2) when $x=-3$ equals 2 which does not satisfy the condition for (2).

What am I doing wrong??

Thank you in advance!

Answer 1

In order to have $f(x)=x|x|+1=-3$, $x$ must be negative (since $x|x|+1\gt0$ if $x\ge0$), in which case we have $|x|=-x$, so that $f(x)=-x^2+1=-3$, or $x^2=4$, and thus $x=-2$ (dismissing the positive square root $x=2$ because we've already concluded that $x$ must be negative).