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.
 A: The original function $f$ takes an $x$ and spits out $y=f(x)$. The inverse is a function $g$ that takes any $y$ from the range of $f$ and spits out $x=g(y)$ such that $g(f(x))=x$ for any $x$ in the domain. So you want to solve for $y$ - you already almost got the correct answer. Once you get $x=g(y)$ for some function $g$, check that $g(f(x)) = x$ to make sure.
A: Let $$y=\frac{1-x^3}{x^3}$$ then $$y=\frac{1}{x^3}-1$$
then we have $$x=\frac{1}{\sqrt[3]{1+y}}$$
A: 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:
$$ x = \frac {1}{\sqrt[3]{1+y}}$$
A: You have almost done, but made a small mistake (me too in the first version of the answer): From 
$$y = \frac{1}{x^3}-1$$ 
you get
$$\frac{1}{x^3} = 1+y$$
and therefore
$$x^3 = \frac{1}{1+y}$$
This show that the inverse is for $-1 < y \le 0$
$$f^{-1}(y) = x = \frac{1}{(1+y)^{\frac{1}{3}}}$$
The range restriction for $y$ comes from the fact, that the range of the function is  $-1 < f(x) \le 0$
