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 June 17th, 2018, 10:58 AM #1 Senior Member   Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 How to solve for f(x) Solve the equation, find the function f(x), where $\displaystyle f^{-1}(x)$ is inverse of f(x), $\displaystyle f(x)=f^{-1}(x)$ Last edited by skipjack; June 17th, 2018 at 10:13 PM.
 June 17th, 2018, 11:53 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 There is no single answer- there are many functions that are there own inverse. One obvious one is f(x)= x. Another is f(x)= 1/x.
 June 17th, 2018, 12:53 PM #3 Senior Member   Joined: Aug 2012 Posts: 2,157 Thanks: 631 $f(x) = -x$, which works in the real or complex numbers. $f(z) = \overline{z}$, the complex conjugate of $z$.
 June 18th, 2018, 12:39 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1967 $\displaystyle f(z) = \frac{a - bz}{b - cz}$, where $a$, $b$ and $c$ are constants that don't satisfy $ac = b^2\!$, and $z$ doesn't satisfy $b - cz = 0$.
 June 19th, 2018, 09:51 AM #5 Senior Member   Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 I post my work below: $\displaystyle f(x)=f^{-1}f^{-1}(x)=f^{-1}(x)$ $\displaystyle f_1 ^{-1}(x)=x=f_1 (x) \; \;$ and $\displaystyle \; \; f_2 ^{-1}(x)=x$ so we got first solution $\displaystyle f_1 (x)=x$ Now I don't understand how to get the second solution $\displaystyle y=\frac{1}{x}$ from $\displaystyle f_2 ^{-1} (x) =x$ Last edited by skipjack; June 19th, 2018 at 10:13 AM.
 June 19th, 2018, 10:39 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,285 Thanks: 1967 Your first line, $f(x)=f^{-1}f^{-1}(x)=f^{-1}(x)$, isn't a consequence of the given information. The given information implies that when the function is nested to an even depth, the result simplifies to $x$, and that when the function is nested to an odd depth, the result simplifies to $f(x)$.
 June 21st, 2018, 04:22 AM #7 Senior Member   Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 I see but let me explain below : Use $\displaystyle f(x)=f^{-1} f^{-1} (x)$ so equation now is $\displaystyle f^{-1}(x)=f^{-1} f^{-1} (x)$ or $\displaystyle f^{-1}(x)=x$ From the system of equations $\displaystyle \begin{cases} f(f^{-1} (x) )=f(x) \\ f(x)=f^{-1} (x) \end{cases} \Rightarrow f(f^{-1} (x))=f^{-1}(x)$ For $\displaystyle f^{-1}(x)=T$ then $\displaystyle f(t)=t$ which is the first solution The second solution $\displaystyle y=1/x$ will take long write so im not posting it
June 21st, 2018, 05:53 AM   #8
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 Originally Posted by idontknow How to solve for f(x) Solve the equation
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