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June 17th, 2018, 09:58 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27  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 09:13 PM. 
June 17th, 2018, 10:53 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
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, 11:53 AM  #3 
Senior Member Joined: Aug 2012 Posts: 2,010 Thanks: 574 
$f(x) = x$, which works in the real or complex numbers. $f(z) = \overline{z}$, the complex conjugate of $z$. 
June 17th, 2018, 11:39 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,529 Thanks: 1750 
$\displaystyle f(z) = \frac{a  bz}{b  cz}$, 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, 08:51 AM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 
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 09:13 AM. 
June 19th, 2018, 09:39 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,529 Thanks: 1750 
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, 03:22 AM  #7 
Senior Member Joined: Dec 2015 From: Earth Posts: 239 Thanks: 27 
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, 04:53 AM  #8 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,127 Thanks: 716 Math Focus: Physics, mathematical modelling, numerical and computational solutions  

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