Inverse Function of $ f(x) = \frac{1-x^3}{x^3} $

Question

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.

Answer 1

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 $-1 < f(x) \le 0$

Answer 2

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.

Answer 3

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:

$$ x = \frac {1}{\sqrt[3]{1+y}}$$

Answer 4

Let $$y=\frac{1-x^3}{x^3}$$ then $$y=\frac{1}{x^3}-1$$ then we have $$x=\frac{1}{\sqrt[3]{1+y}}$$