Author Topic: Inverse (Read 3023 times) Tweet Share

TrueTears · « **on:** March 19, 2009, 05:40:51 pm »

Show that $a \ne d$ if $f(x) = f^{-1}(x) = \frac{ax+b}{cx+d}$ and $abcd \ne 0$

I've tried long division $\frac{ax+b}{cx+d}$ then found the $f^{-1}(x)$ but I don't see how I can show $a \ne d$

Any help would be much appreciated

Thanks!

EDIT: omg i'm an idiot, sorry guys its meant to be $a \ne d$ (not b) -_-

pHysiX · « **Reply #1 on:** March 19, 2009, 09:07:49 pm »

hmm is it me or is latex down?...lol i can't read it >.<

TrueTears · « **Reply #2 on:** March 19, 2009, 09:08:50 pm »

Quote from: pHysiX on March 19, 2009, 09:07:49 pm

hmm is it me or is latex down?...lol i can't read it >.<

nope i can see LaTeX o.o

lacoste · « **Reply #3 on:** March 19, 2009, 09:21:20 pm »

latex is readable, the problem is that i have no idea of inverse.

pHysiX · « **Reply #4 on:** March 19, 2009, 09:29:08 pm »

ok so here's wat i came up with...but not sure how correct i am:

f(x)= ax+b/cx+d
find f^-1 by doing the normal stuff

then get that answer and make it equal to ax+b/cx+d
and then multiply the denominators up so u don't have fractions.
expand them, you will find that they're quadratics.
following the rule of equating coefficients:
we see that a=-d
since a,b,c,d are never zero, hence, a is never equal to d

i'll latex this after i fix up my firefox >.<
=]

TrueTears · « **Reply #5 on:** March 19, 2009, 09:31:35 pm »

Quote from: pHysiX on March 19, 2009, 09:29:08 pm

ok so here's wat i came up with...but not sure how correct i am:

f(x)= ax+b/cx+d
find f^-1 by doing the normal stuff

then get that answer and make it equal to ax+b/cx+d
and then multiply the denominators up so u don't have fractions.
expand them, you will find that they're quadratics.
following the rule of equating coefficients:
we see that a=-d
since a,b,c,d are never zero, hence, a is never equal to d

i'll latex this after i fix up my firefox >.<
=]

oh great, thanks physix that works perfectly.

Thanks again!

pHysiX · « **Reply #6 on:** March 19, 2009, 09:44:20 pm »

i know this is the wrong place, but is anyone using firefox? i don't know y my latex is not working right now >.<

Flaming_Arrow · « **Reply #7 on:** March 19, 2009, 09:45:45 pm »

Quote from: pHysiX on March 19, 2009, 09:44:20 pm

i know this is the wrong place, but is anyone using firefox? i don't know y my latex is not working right now >.<

yer im using it , it works for me

TrueTears · « **Reply #8 on:** March 19, 2009, 09:46:56 pm »

Quote from: pHysiX on March 19, 2009, 09:44:20 pm

i know this is the wrong place, but is anyone using firefox? i don't know y my latex is not working right now >.<

same here im using it

pHysiX · « **Reply #9 on:** March 19, 2009, 09:51:02 pm »

dammit...i changed to IE and still not working...anything special i need to install to make it work?

TrueTears · « **Reply #10 on:** March 19, 2009, 10:03:43 pm »

Have you tried restarting computer

lol I'm no expert at computers -_-

shinny · « **Reply #11 on:** March 19, 2009, 10:15:06 pm »

You shouldn't need to install anything since they're essentially just image files. Try refreshing the page a couple of times...I had a similar problem yesterday which was fixed by refreshing once or twice.

Mao · « **Reply #12 on:** March 19, 2009, 10:40:47 pm »

Another method:

$f( f^{-1} (x) ) = x$

$\implies \frac{a \frac{ax+b}{cx+d} +b }{ c \frac{ax+b}{cx+d} + d} = x$ , this must be the case for $x\in \mathbb{R}\backslash \left\{ \frac{-d}{c} \right\}$

$\implies \frac{a(ax+b) + b(cx+d)}{cx+d} = x \cdot \frac{c(ax+b)+d(cx+d)}{cx+d}$

$\implies (a^2 + bc)x + (ab + bd) = (ac + dc)x^2 + (bc +d^2)x$

$\implies -c (a+d)x^2 + (a^2-d^2) x + b(a+d) = 0$

$\implies (a+d) \left( -cx^2 + (a-d)x + b \right) = 0$

this can only be true if $a+d=0$ or $-cx^2 + (a-d)x + b = 0$

since the above is true for [ $x\in \mathbb{R}\backslash \left\{ \frac{-d}{c} \right\}$ , the latter coefficients in the quadratic expression must be zero, i.e. $c = 0$ , $b=0$ , $a = d$ , however, $abcd \neq 0$ , this implies a,b,c nor d are zero, hence the quadratic expression does not satisfy the domain.

hence $a=-d$

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TrueTears

Inverse

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