Functions inverse + domain

Question

The function $f(x)$ is defined by $f(x)=2x^2-3\quad\{x\in\mathbb R,x<0\}$. Determine

(a) $f^{-1}(x)$ clearly stating its domain

(b) the values of $a$ for which $f(a)=f^{-1}(a)$.

Answer for part (a):

The domain of $f^{-1}(x)$ is the range of $f(x)$.

$f(x)=2x^2-3\quad\{x\in\mathbb R,x<0\}$ has range $f(x)>-3$

Hence $f^{-1}(x)$ must be the negative square root

$f^{-1}=-\displaystyle\sqrt\frac{x+3}2$ has domain $x\in\mathbb R, x>-3$

Can anyone explain why the inverse must be the negative square root?

Answer 1

To make $f$ invertible it was defined for $x<0$. Then the inverse function will be the negative square root, since that will pick the value in your original domain $x<0$. You could have just as well picked $x>0$ to define the domain of $f$, and then you would take the positive square root.

Answer 2

remember that $$domain(f^{-1}) = range(f), \quad range(f^{-1}) = domain(f).$$ the function $f(x)= y = 2x^2 - 3$ is $1-1$ on the domain $(-\infty, 0)$ which becomes the range of $f^{-1}(x) = -\sqrt[3]{\frac{x+3}2}.$