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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.

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

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  • $\begingroup$ Why $ 0 \leq y \lt 1 $? im missing simple math basics here. i'm sorry. $\endgroup$ – loop Sep 12 '13 at 11:50
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    $\begingroup$ Because the range of $f$ is $0 \le f(x) < 1$. $\endgroup$ – gammatester Sep 12 '13 at 11:53
  • $\begingroup$ yeah. now i got that. thanks! $\endgroup$ – loop Sep 12 '13 at 11:56
  • $\begingroup$ @gammatester you missed the OP function. it seems to be the answer of another question! $\endgroup$ – RSh Sep 12 '13 at 11:58
  • $\begingroup$ You're right @I'mtoo, I have made the corrections. $\endgroup$ – gammatester Sep 12 '13 at 12:38
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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.

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  • $\begingroup$ yeah but how do I proceed when on my left hand i have $ \frac{1}{x^3} $ How do I solve this to $ x = $ without multiplying on both sides with $ x^3 $ $\endgroup$ – loop Sep 12 '13 at 11:47
  • $\begingroup$ @DominiqueLüber: I give you a complete argument. $\endgroup$ – RSh Sep 12 '13 at 11:49
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Let $$y=\frac{1-x^3}{x^3}$$ then $$y=\frac{1}{x^3}-1$$ then we have $$x=\frac{1}{\sqrt[3]{1+y}}$$

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  • $\begingroup$ is $ 1/a = b => a = 1/b $ ? Or how do you came up with that? $\endgroup$ – loop Sep 12 '13 at 11:48
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    $\begingroup$ absolutely yes! if $y+1=\frac{1}{x^3}$ then $x^3=\frac{1}{y+1}$ and we have the latter equality! (you must implicitly not $a=0$ and not $b=0$) $\endgroup$ – RSh Sep 12 '13 at 11:52
  • $\begingroup$ of course. to early in the day. I was blind. :) thanks! $\endgroup$ – loop Sep 12 '13 at 11:54
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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:

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

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  • $\begingroup$ Oh well, no points for me :P $\endgroup$ – Crex Sep 12 '13 at 11:59

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