find the special function in this problem

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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Hint: $y=f^{-1}(x)=\frac{x}{x-1}$. – Tapu Oct 11 '11 at 23:59
@Swapan - may you explain how do you get your hint? – Victor Oct 12 '11 at 0:01
FYI "special functions" has a very specific meaning in mathematics. There are several monographs filled with curious results about the properties of these functions. For example this. – kahen Oct 12 '11 at 0:02
@Victor, I have edited to start from $y=f(x)\Rightarrow x=f^{-1}(y)$. – Tapu Oct 12 '11 at 0:08

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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WAIT. If $f = f^{-1}$, then $f^{-1}\circ f^{-1}$ is the identity, not $f^{-1}$ again... – GEdgar Oct 12 '11 at 0:08
@GEdgar you're right, sorry. I was thinking of f inverted twice instead of f inverse applied to itself – Paulpro Oct 12 '11 at 0:11
@GEdgar - what do you mean and how do you get that? – Victor Oct 12 '11 at 0:41
@Victor: Try computing it yourself! – GEdgar Oct 12 '11 at 1:43

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

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

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