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Let $f(x)=\frac{\alpha x}{x+1},x\neq-1$. Then for what value of $\alpha$ is $f(f(x))=x$?

My approach:
Given that,
$f(f(x))=x$
Applying $f^{-1}$ on both sides, we get
$f(x)=f^{-1}(x)$
The above condition is only true when the curve $y=f(x)$ intersects with $y=x$.

$\therefore f(x)=x\implies \frac{\alpha x}{x+1}=x$
On solving this, we get $\alpha $ as a function of $x$.

I know that there is another method to solve this in which we find the expression for $f(f(x))$ and equate it to $x$.

I don't need that solution. I just want to know what is wrong with this approach as this was the first thing that came to my mind on seeing the question and the method was much shorter compared to the above-mentioned method.

Anybody with alternative methods are welcome to share.

Thanks in advance.

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  • $\begingroup$ Is your result wrong that you know of? (Did your source give a solution?) $\endgroup$ – coffeemath Oct 18 '18 at 8:37
  • $\begingroup$ That's not the only solution. Consider $f(x)=k-x$ and $f(x)=k/x$. $\endgroup$ – TheSimpliFire Oct 18 '18 at 8:40
  • $\begingroup$ @coffeemath Yeah, as per the source, the answer is -1. They have used the other method that I have specified above or what gimusi has done. $\endgroup$ – Vaibhav Oct 18 '18 at 8:46
  • $\begingroup$ @TheSimpliFire I see. will delete comment. $\endgroup$ – coffeemath Oct 18 '18 at 8:51
  • $\begingroup$ @TheSimpliFire How? I don't understand. $\endgroup$ – Vaibhav Oct 18 '18 at 9:32
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In general, the solutions of $f(f(x))=x$ are indeed the same as those of $f(x)=f^{-1}(x)=x$,

$$\frac{\alpha x}{x+1}=\frac x{\alpha-x}=x.$$

For a given $\alpha\ne-1$, there are two solution points (intersections of equilateral hyperbolas and a straight line).

But when $\alpha=-1$, $f(x)=f^{-1}(x)$ becomes an identity, i.e. it holds whatever $x$, and in general

$$-\frac{x}{x+1}=\frac x{-1-x}\ne x.$$


Below, the solutions of $f(x)=f^{-1}(x)$ for a few $\alpha$. The hyperbolas intersect in pairs along the green diagonal. And for $\alpha=-1$, the hyperbolas merge into one.

enter image description here

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  • 2
    $\begingroup$ So your mistake resides in "The above condition is only true when...". $\endgroup$ – Yves Daoust Oct 18 '18 at 9:30
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Note that when $\alpha = -1$, the vertical and horizontal asymptotes will be equal: $x=-1, y=-1$ and $f(x)=f^{-1}(x)$.

In fact, that is the case for any hyperbola with equal asymptotes:

$$f(x)=\frac{ax+b}{c x+d}, \left(-\frac{d}{c}=\frac{a}{c}\right) \Rightarrow \\ y=\frac{ax+b}{c x+d} \Rightarrow x=\frac{b-yd}{yc-a} \Rightarrow \\ f^{-1}(x)=\frac{b-xd}{xc-a}=\frac{ax+b}{cx+d}.$$ You can equate the original function to its inverse to see the solution $\alpha=-1$ clearly: $$f(x)=f^{-1}(x) \Rightarrow \frac{\alpha x}{x+1}=\frac{x}{\alpha-x} \Rightarrow \\ \frac{\alpha^2x-\alpha x^2-x^2-x}{(x+1)(\alpha -x)}=0 \Rightarrow \\ \frac{(\alpha +1)(\alpha-x-1)x}{(x+1)(\alpha -x)}=0$$

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  • $\begingroup$ The denominator can be ignored when the right side is $0$. But why is the denominator still there anyway? Didn't you multiply it out? $\endgroup$ – mr_e_man Oct 18 '18 at 10:19
  • $\begingroup$ No, I moved the RHS to the LHS and simplified. It is a way of mentioning that $x\ne -1$. Otherwise, the denominator can be omitted. $\endgroup$ – farruhota Oct 18 '18 at 10:23
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$f(f(x))=\alpha^2x/((x+1)(\alpha x/(x+1)+1))$ The equation you have to solve is: $$f(f(x))=x$$ So you get: $$\alpha=-1$$ and $$\alpha=x+1$$

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