F(x) coincides with its inverse function

[alexmahone]

Suppose that $\displaystyle f(x)=\frac{ax+b}{cx+d}$. What conditions on $\displaystyle a,\ b,\ c,\ d$ are necessary and sufficient in order that $\displaystyle f(x)$ coincide with its inverse function.

My attempt:

$\displaystyle f(f(x))=\frac{a\left(\frac{ax+b}{cx+d}\right)+b}{c\left(\frac{ax+b}{cx+d}\right)+d}=\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=\frac{(a^2+bc)x+ab+bd}{(ac+cd)x+bc+d^2}$

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

$\displaystyle \implies f(f(x))=x$

$\displaystyle \implies\frac{(a^2+bc)x+ab+bd}{(ac+cd)x+bc+d^2}=x$

$\displaystyle \implies(a^2+bc)x+ab+bd=(ac+cd)x^2+(bc+d^2)x$

$\displaystyle \implies c(a+d)x^2+(d^2-a^2)x-b(a+d)=0$

$\displaystyle \implies (a+d)[cx^2+(d-a)x-b]=0$

$\displaystyle a+d=0$ or $\displaystyle b=c=0$, $\displaystyle a=d$

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Is that correct?

 

[earboth]

Suppose that $\displaystyle f(x)=\frac{ax+b}{cx+d}$. What conditions on $\displaystyle a,\ b,\ c,\ d$ are necessary and sufficient in order that $\displaystyle f(x)$ coincide with its inverse function.

My attempt:

$\displaystyle f(f(x))=\frac{a\left(\frac{ax+b}{cx+d}\right)+b}{c\left(\frac{ax+b}{cx+d}\right)+d}=\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=\frac{(a^2+bc)x+ab+bd}{(ac+cd)x+bc+d^2}$

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

$\displaystyle \implies f(f(x))=x$

$\displaystyle \implies\frac{(a^2+bc)x+ab+bd}{(ac+cd)x+bc+d^2}=x$

$\displaystyle \implies(a^2+bc)x+ab+bd=(ac+cd)x^2+(bc+d^2)x$

$\displaystyle \implies c(a+d)x^2+(d^2-a^2)x-b(a+d)=0$

$\displaystyle \implies (a+d)[cx^2+(d-a)x-b]=0$

$\displaystyle a+d=0$ or $\displaystyle b=c=0$, $\displaystyle a=d$

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Is that correct? <--- yes

I've compared the co-efficients at:

$ \displaystyle f(x)=\frac{ax+b}{cx+d}$ ... and... $\displaystyle f^{-1}(x)=\frac{-dx+b}{cx-a} $

which yields your results.