Inverse Function of $ f(x) = \frac{1-x^3}{x^3} $ I need to show that these function has a continuous inverse function and find this inverse function.
$$ f(x) = \frac{1-x^3}{x^3} $$
Defined on $ (1,\infty) $
I think I need to check for bijectivity. Don't know how.
I tried to solve the function to $x$ then. But somehow I only end up with $ -\frac{1}{x^3} = -1 -y $ and don't know how to get to $x = ...$
Maybe there is no inverse function!? Or maybe just on the defined area? I don't know.

So, you know how to get x³ = ..., right? If so, just do the cube root:
$$x = \sqrt[3]{ \frac{1}{1+y} }$$
You can simplify this to become:
$$ x = \frac {1}{\sqrt[3]{1+y}}$$
