Find the inverse, domain and range of $f(x)=\frac{1}{\sqrt{-2x}}$

Question

The inverse I am getting is $f^{-1}(x)= \frac{1}{2x^{2}}$. The domain and range of $f(x)$ is $x<0$ , $y>0$. The domain and range of $f^{-1}(x)$ is $x>0$ , $y>0$ though.

What am I doing wrong? Is my inverse function incorrect?

Answer 1

Starting for $x < 0$ with $$ y = \frac{1}{\sqrt{-2x}} > 0 \Rightarrow \\ y^2 = -\frac{1}{2x} > 0 \Rightarrow \\ x = - \frac{1}{2y^2} < 0 $$ we get the inverse function $$ f(x) = -\frac{1}{2x^2} $$ with the domain $(0, \infty)$ and range $(0, -\infty)$.