A family of functions where each member is its own inverse?

Mar 2014
42
0
Canada
A family of functions is a set of functions that share one or more properties. ie: The family of quadratics with zeros 1 and 10, or the linear functions with a slope of 20.

there is a family of linear functions where each member is its own inverse. What linear property defines the family?

(I don't really know how to start this, though I think i understand what the question is asking. Help would be appreciated!)
 

HallsofIvy

MHF Helper
Apr 2005
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I don't know what you mean by "linear property". Of course, if "f is its own inverse" then \(\displaystyle f= f^{-1}\) or \(\displaystyle f(f(x))= x\).
 
Mar 2014
42
0
Canada
Linear property is a property of the line, like the slope or the zeros.
 

romsek

MHF Helper
Nov 2013
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California
Linear property is a property of the line, like the slope or the zeros.
solving $f(f(x))=x$ for the parameters of slope m, and intercept b, quickly shows that

$f(x)=mx+b$

$f(f(x))=m(mx+b)+b=x$

$m^2x+bm+b = x$

$(m^2=1) \wedge (b(m+1)=0)$

$m=-1$ and $b$ can take any value.

so $f(x)=b-x$

a quick check shows

$f(f(x))=b - (b-x) = (b-b) + x = x$
 
Mar 2014
42
0
Canada
solving $f(f(x))=x$ for the parameters of slope m, and intercept b, quickly shows that

$f(x)=mx+b$

$f(f(x))=m(mx+b)+b=x$

$m^2x+bm+b = x$

$(m^2=1) \wedge (b(m+1)=0)$

$m=-1$ and $b$ can take any value.

so $f(x)=b-x$

a quick check shows

$f(f(x))=b - (b-x) = (b-b) + x = x$
I don't really understand. This seems pretty complex, since I'm in grade 11 and this question would be at that level. Is there an easier way to prove it? Or at least to explain it in more detail?

Thanks by the way!
 

romsek

MHF Helper
Nov 2013
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if you know what linear means then what I posted isn't too complex for you to understand. It's just algebra.

maybe the $\wedge$ symbol confused you. It just means AND as in $m^2=1$ AND $b(m+1)=0$

Just read through it, it's not that hard.
 
Jun 2009
675
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(Without the fancy notation), you are told that the relationship is to be linear, so start with \(\displaystyle y=ax+b\).
Inverting, (solving for x in terms of y) gets you
\(\displaystyle x=\frac{1}{a}y-\frac{b}{a}.\)
and switching the variables
\(\displaystyle y=\frac{1}{a}x-\frac{b}{a}.\)
which is the inverse relationship.
For this to be the same as the original, we need \(\displaystyle a\) to be the same as \(\displaystyle 1/a\) implying that \(\displaystyle a^{2}=1, \text{ so that } a=\pm 1\), and \(\displaystyle b=-b/a\) implying that \(\displaystyle a=-1,\) with \(\displaystyle b\) being able to take any value.

That gets you \(\displaystyle y=b-x\), (from which \(\displaystyle x=b-y).\)
 

SlipEternal

MHF Helper
Nov 2010
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The method they are using is equating coefficients. The only way $ax+b = \dfrac{1}{a}x - \dfrac{b}{a}$ is if $a = \dfrac{1}{a}$ and $b = -\dfrac{b}{a}$ since the coefficient of $x$ is $a$ in the first equation and $\dfrac{1}{a}$ in the second, and the coefficient of 1 is $b$ in the first equation and $-\dfrac{b}{a}$ in the second.

Note that $y=x$ is its own inverse, as well (using BobP's notation, that is when $a=1$ and $b=0$, and using romsek's notation, that is when $m=1, b=0$).
 
Mar 2014
42
0
Canada
(Without the fancy notation), you are told that the relationship is to be linear, so start with \(\displaystyle y=ax+b\).
Inverting, (solving for x in terms of y) gets you
\(\displaystyle x=\frac{1}{a}y-\frac{b}{a}.\)
and switching the variables
\(\displaystyle y=\frac{1}{a}x-\frac{b}{a}.\)
which is the inverse relationship.
For this to be the same as the original, we need \(\displaystyle a\) to be the same as \(\displaystyle 1/a\) implying that \(\displaystyle a^{2}=1, \text{ so that } a=\pm 1\), and \(\displaystyle b=-b/a\) implying that \(\displaystyle a=-1,\) with \(\displaystyle b\) being able to take any value.

That gets you \(\displaystyle y=b-x\), (from which \(\displaystyle x=b-y).\)
Oh I see. That helps a lot!

Though, I'm not really sure I understand that last bit. Why does \(\displaystyle b=-b/a\)? And why does that imply that \(\displaystyle a=-1\), with \(\displaystyle b\) taking any value?

Thanks by the way, everyone. :)
 

SlipEternal

MHF Helper
Nov 2010
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Oh I see. That helps a lot!

Though, I'm not really sure I understand that last bit. Why does \(\displaystyle b=-b/a\)? And why does that imply that \(\displaystyle a=-1\), with \(\displaystyle b\) taking any value?

Thanks by the way, everyone. :)
Read my response directly above your question. It explains exactly why you need $a$ to be equal to $\dfrac{1}{a}$ and $b = -\dfrac{b}{a}$. It is the method of equating coefficients. You have two equations for $y$. Those equations are both equal. So, the coefficient of $x$ must be the same, and the coefficient of $1$ must be the same in both equations.

As for why $b = -\dfrac{b}{a}$ implies $a=-1$ and $b$ may take any value, if $b \neq 0$, then you may divide both sides by $b$. Hence, $a = -1$. If $b=0$, then $a$ may take any value that satisfies $a^2=1$.