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If $f(x)=\frac{x}{x-1},$ what is $f^{-1}(f^{-1}(x))$ and $f^{-1}(f^{-1}(f^{-1}(x)))$?

Please help to evaluate if you know. This confuse me a lot! Thanks in advance. is those able to express in term of $x$?

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    @Victor, I have edited to start from $y=f(x)\Rightarrow x=f^{-1}(y)$.2011-10-12

3 Answers 3

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The inverse of that function is itself

$f(x) = f^{-1}(x)$

Derivation:

Find the inverse by replace $f(x)$ with $x$ and all $x$'s with $f^{-1}(x)$, then solving for $f^{-1}(x)$

$f(x) = x/(x - 1)$ $x = f^{-1}(x)/(f^{-1}(x) - 1)$ $x*(f^{-1}(x) - 1) = f^{-1}(x)$ $x*f^{-1}(x) - x = f^{-1}(x)$ $x*f^{-1}(x) = f^{-1}(x) + x$ $x*f^{-1}(x) - f^{-1}(x) = x$ $f^{-1}(x)(x - 1) = x$

$f^{-1}(x) = x/(x - 1)$

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    @Victor: Try computing it yourself!2011-10-12
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It’s easy enough to solve for $f^{-1}$, as PaulPRO did in his answer, but in this case it’s also fairly easy to see intuitively why $f$ is its own inverse. First divide it out: $f(x) = \frac{x}{x-1} = 1 + \frac1{x-1}.$ To calculate $f(x)$ from $x$ you first subtract $1$ to get $x-1$. Then you take the reciprocal to get $\dfrac1{x-1}$. And finally you add $1$ to get $1+\dfrac1{x-1}$.

Now what happens when you try to reverse the action of $f$? You have to subtract $1$, take the reciprocal, and add $1$, which is exactly what $f$ did in the first place. Thus, $f^{-1}$ must be the same function as $f$.

To make this a bit less informal, notice that what we’ve actually done is split $f$ up as a composition of three functions, which can be pictured as follows: $x \stackrel{g}\longmapsto x-1 \stackrel{h}\longmapsto \frac1{x-1} \stackrel{j}\longmapsto 1+\frac1{x-1},$ with $f=j\circ h\circ g$.

Clearly $g$ (subtracting $1$) and $j$ (adding $1$) are inverses of each other, so $j=g^{-1}$, $f=g^{-1}\circ h \circ g$, and we can improve the picture: $x \stackrel{g}\longmapsto x-1 \stackrel{h}\longmapsto \frac1{x-1} \stackrel{g^{-1}}\longmapsto 1+\frac1{x-1}.\tag{1}$ Note that $h$ is its own inverse: if you take the reciprocal twice, you’re back where you started.

It’s a basic fact that if the functions involved are invertible, $(\phi\circ\psi)^{-1}=\psi^{-1}\circ\phi^{-1}$, so $f^{-1}=$ $(g^{-1}\circ h\circ g)^{-1} = g^{-1}\circ h^{-1} \circ (g^{-1})^{-1} = g^{-1}\circ h \circ g = f$. In more detail, $f^{-1}$ must reverse the chain of transformations shown in $(1)$: $1+\frac1{x-1} \stackrel{(g^{-1})^{-1}}\longmapsto \frac1{x-1} \stackrel{h^{-1}}\longmapsto x-1 \stackrel{g^{-1}}\longmapsto x.$ Since $(g^{-1})^{-1}=g$ and $h^{-1}=h$, this is simply $1+\frac1{x-1} \stackrel{g}\longmapsto \frac1{x-1} \stackrel{h}\longmapsto x-1 \stackrel{g^{-1}}\longmapsto x,$ with exactly the same composition of simpler functions as in $(1)$.

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Part 1 - Finding the inverse function

It may help you to look at the definition and some examples at here: (inverse function link-1

1-replace f(x) with y. This is done to make the rest of the process easier.

$y=\frac{x}{x-1},$

2-Replace every x with a y and replace every y with an x.

$x=\frac{y}{y-1},$

3-Solve the equation from Step 2 for y.

$y=\frac{x}{x-1},$ Note that this is the same as the original f(x), but this is not always the case.

4-Replace y with $f^{-1}(x)$ In other words, we’ve managed to find the inverse at this point!

$f^{-1}(x)=\frac{x}{x-1},$

Part 2 - Finding $f^{-1}(f^{-1}(f^{-1}(x)))$

to find $f^{-1}(f^{-1}(f^{-1}(x)))$, we find $f^{-1}(z)$ where $z=f^{-1}(f^{-1}(x))$

using $f^{-1}(x)=\frac{x}{x-1},$ replace the value of x by $x/(x-1)$

$z=(f^{-1}(f^{-1}(x)) = ([x/(x-1)])/([x/(x-1)]-1)$

simplifying, you get:

$z=x$

now $f^{-1}(z)$ = $f^{-1}(x)$ which is = $\frac{x}{x-1},$

so

$f^{-1}(f^{-1}(f^{-1}(x)))=\frac{x}{x-1}$