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Question Number 1278 by Rasheed Soomro last updated on 19/Jul/15
Determine f(x) when f( f(x))=x  .
$${Determine}\:{f}\left({x}\right)\:{when}\:{f}\left(\:{f}\left({x}\right)\right)={x}\:\:. \\ $$
Commented by prakash jain last updated on 19/Jul/15
f(x)=((ax+b)/(cx+d))  f(f(x))=((a((ax+b)/(cx+d))+b)/(c((ax+b)/(cx+d))+d))=x  a^2 x+ab+bcx+bd=c(a+d)x^2 +(bc+d^2 )x  Comparing coefficients  a+d=0  f(x)=((ax+b)/(cx−a))  f(f(x))=((a((ax+b)/(cx−a))+b)/(c((ax+b)/(cx−a))−a))=((a^2 x+ab+bcx−ab)/(acx+bc−acx+a^2 ))=x  More solution can be found.
$${f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$${f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}+{d}}+{b}}{{c}\frac{{ax}+{b}}{{cx}+{d}}+{d}}={x} \\ $$$${a}^{\mathrm{2}} {x}+{ab}+{bcx}+{bd}={c}\left({a}+{d}\right){x}^{\mathrm{2}} +\left({bc}+{d}^{\mathrm{2}} \right){x} \\ $$$$\mathrm{Comparing}\:\mathrm{coefficients} \\ $$$${a}+{d}=\mathrm{0} \\ $$$${f}\left({x}\right)=\frac{{ax}+{b}}{{cx}−{a}} \\ $$$${f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}−{a}}+{b}}{{c}\frac{{ax}+{b}}{{cx}−{a}}−{a}}=\frac{{a}^{\mathrm{2}} {x}+{ab}+{bcx}−{ab}}{{acx}+{bc}−{acx}+{a}^{\mathrm{2}} }={x} \\ $$$$\mathrm{More}\:\mathrm{solution}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}. \\ $$
Commented by prakash jain last updated on 19/Jul/15
f(x)=x  f(x)=k−x  f(x)=(k/x)  Also more solutions are possible
$${f}\left({x}\right)={x} \\ $$$${f}\left({x}\right)={k}−{x} \\ $$$${f}\left({x}\right)=\frac{{k}}{{x}} \\ $$$$\mathrm{Also}\:\mathrm{more}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{possible} \\ $$
Answered by prakash jain last updated on 19/Jul/15
Some solutions are given in comments  but more solutions exist.
$$\mathrm{Some}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{given}\:\mathrm{in}\:\mathrm{comments} \\ $$$$\mathrm{but}\:\mathrm{more}\:\mathrm{solutions}\:\mathrm{exist}. \\ $$

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