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From my text book it says that $f(x)= x^3$ and $f^{-1}(x) = \sqrt[3]{x}$ , which I totally agree with.

why does $f(x)= \frac 1 {x-1}$ and $f^{-1}(x)= \frac 1 {x + 1}$ and not equal $f^{-1}(x)= \frac 1 {x+1}$?

I know when you inverse a function you reverse the sign values. Can anyone explain this to me a little more thoroughly?

(use curl brackets for multichar objects in FRAC)

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up vote 1 down vote accepted

I'm not sure what you mean by "you reverse the sign values" - generally you cannot just "tweak" the function slightly in order to get the inverse. If $f(x)=1/(x-1)$ then write down $y=1/(x-1)$, swap the $x$ and $y$, and solve for $y$: $$ x=\frac{1}{y-1} \\ x(y-1)=1 \\ xy=x+1 \\ y=\frac{x+1}{x}=1+\frac{1}{x} $$

You can check your answer by verifying the following: $$f(f^{-1}(x))=x \quad\mbox{and}\quad f^{-1}(f(x))=x$$

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Thank you for showing the work on this. From the last part, y= (x+1)/x how does that equal 1 + 1/x? That last step doesn't seem to quit click with me. – Tyler Zika Nov 5 '12 at 4:08
@TylerZika: Recall that a fraction $\frac{a+b}{c}$ simplifies to $\frac{a}{c}+\frac{b}{c}$, so $\frac{x+1}{x}=\frac{x}{x}+\frac{1}{x}=1+\frac{1}{x}$. – wj32 Nov 5 '12 at 4:10
@TylerZika $$\dfrac{(x+1)}{x} = \dfrac{x}{x} + \dfrac1x = 1 + \dfrac1x$$ – user17762 Nov 5 '12 at 4:10
@wj32 Awesome, thank you! – Tyler Zika Nov 5 '12 at 5:16
@Marvis why does 2y= x-3 equal y = 1/2(x-3) and not (x-3)/2 with that logic? – Tyler Zika Nov 5 '12 at 5:56

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