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December 23rd, 2015, 03:38 AM  #1 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0  Selfinverse function
A function is said to be a selfinverse if f(x)=f^1(x) for all x in the domain. a) Show that f(x)=1/x is a selfinverse function. b) Find the value of the constant k so that g(x)=(3x5)/(x+k) is a selfinverse function. A) How should I prove that: y=1/x yx=1 x=1/y f^1(x)=1/x But in this way it does not make sense. B) ?? Last edited by skipjack; December 23rd, 2015 at 03:44 AM. 
December 23rd, 2015, 03:43 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
(a) If y = 1/x, x = 1/y, so the inverse of f(x) = 1/x is f$\,^{1}$(x) = 1/x. If you don't understand that, what exactly is puzzling you? Last edited by skipjack; December 23rd, 2015 at 03:46 AM. 
December 23rd, 2015, 03:55 AM  #3 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0 
I just wanted to check for the first one, because it seemed extremely easy, but it carries the highest amount of points. I thought I was wrong but now it seems I am right. The second problem. How should I find the k? Last edited by skipjack; December 23rd, 2015 at 04:15 AM. 
December 23rd, 2015, 04:20 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
(b) Solving x = (3g  5)/(g + k) for g gives g = (kx  5)/(x  3), so k = 3 for g to be selfinverse.

December 23rd, 2015, 04:41 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
Formally, you find solution for $k$ that are independent of $x$ to $$\begin{aligned} {3x+5 \over x+k} &= {kx+5 \over x3} \\ (3x+5)(x3) &= (kx5)(x+k) \\ 3x^2 4x 15 &= kx^2 (k^2 5)x +5k \\ (3+k)x^2+(k^29)x(15+5k) &= 0 \end{aligned}$$ The only way this can be true for all $x$ (independent of $x$) is if all the coefficients are zero. 
December 23rd, 2015, 05:24 AM  #6 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0 
I understand thank you
Last edited by med1student; December 23rd, 2015 at 05:33 AM. 
December 23rd, 2015, 05:46 AM  #7 
Member Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0 
It should be kx5,not kx+5

December 23rd, 2015, 05:49 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,094 Thanks: 846 
The crucial point of "inverse functions" is that $\displaystyle f(f^{1}(x))= x$ and $\displaystyle f^{1}(f(x))= x$. So a function, f, is "self inverse" if and only if $\displaystyle f(f(x))= x$. If $\displaystyle f(x)=\frac{1}{x}$ then $\displaystyle f(f(x))= \frac{1}{\frac{1}{x}}= \left(1\right)\left(\frac{x}{1}\right)= x$ and we are done. $\displaystyle g(x)= \frac{3x 5}{x+ k}$ then g is "self inverse" if and only if $\displaystyle g(g(x))= \frac{3\left(\frac{3x 5}{x+ k}\right) 5}{\frac{3x 5}{x+ k}+ k}= \frac{\frac{9x 15}{x+ k} 5}{\frac{3x 5}{x+ k}+ k}$ Now multiply both numerator and denominator by x+ k to get $\displaystyle g(g(x))= \frac{(9x 15) (5x+ 5k}{(3x 5)+ kx+ k^2}= \frac{4x 15 5k}{(3+ k)x+ k^2 5}$. That will be equal to x if and only if $\displaystyle 4x 15 5k= (3+ k)x^2+ (k^2 5)x$. Since the left side has no "$\displaystyle x^2$", in order for these to be equal for all x, the right side must not either we must have 3+ k= 0 or k= 3. In that case, 15 5k= 15 15= 0 also so we have 4x= (9 5)x= 4x. 
December 23rd, 2015, 02:21 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
Followup question: if f(x) = (ax + b)/(cx  a) is a selfinverse function, what condition must be satisfied by the coefficients a, b and c?


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