My Math Forum Self-inverse function

 Algebra Pre-Algebra and Basic Algebra Math Forum

 December 23rd, 2015, 03:38 AM #1 Member   Joined: Aug 2015 From: Podgorica Posts: 61 Thanks: 0 Self-inverse function A function is said to be a self-inverse if f(x)=f^-1(x) for all x in the domain. a) Show that f(x)=1/x is a self-inverse function. b) Find the value of the constant k so that g(x)=(3x-5)/(x+k) is a self-inverse 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: 19,065 Thanks: 1621 (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? Thanks from med1student 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: 19,065 Thanks: 1621 (b) Solving x = (3g - 5)/(g + k) for g gives g = (-kx - 5)/(x - 3), so k = -3 for g to be self-inverse.
 December 23rd, 2015, 04:41 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 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 x-3} \\ (3x+5)(x-3) &= -(kx-5)(x+k) \\ 3x^2 -4x -15 &= -kx^2 -(k^2 -5)x +5k \\ (3+k)x^2+(k^2-9)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. Thanks from med1student
 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 -kx-5,not -kx+5
 December 23rd, 2015, 05:49 AM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,198 Thanks: 872 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: 19,065 Thanks: 1621 Follow-up question: if f(x) = (ax + b)/(cx - a) is a self-inverse function, what condition must be satisfied by the coefficients a, b and c?

 Tags function, selfinverse

,
,
,

,

,

,

,

,

,

,

,

,

,

,

# show that f(x)=(11x 2)/(7x-11) is a self-inverse function

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post WWRtelescoping Calculus 2 September 2nd, 2014 04:57 AM deSitter Algebra 4 April 10th, 2013 01:17 PM rayman Real Analysis 3 February 24th, 2013 08:17 AM jaredbeach Algebra 1 November 17th, 2011 11:58 AM Tear_Grant Calculus 1 April 19th, 2009 03:59 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top