# Thread:

1. ## Constants for Composition of functions

Let f(x) = ax+b and g(x) =cx+d where a,b,c,d are constants. Determine for which constants a,b,c,d it is true that f o g = g o f.

I am not very sure how to approach this problem. I simplifed the f o g and g o f but after that I am not sure how to go from there.

Here is what I did first

f(g(x)) = g(f(x))
a(cx+d) +b = c(ax+b) +d
acx+ad+b = cax + cb +d

then after that I got stuck. I tried switching around the varibles on either side and tried to get them to equal to eachother, however that didn't work. Am I'm going in the right direction or am I totally off? Thank you for your help!!!

acx-acx = cb +d
ad+b = cb +d
ad -d = cb -b
d(a-1) = b(c-1)

that is where I got stuck

2. Originally Posted by hotmail590
Let f(x) = ax+b and g(x) =cx+d where a,b,c,d are constants. Determine for which constants a,b,c,d it is true that f o g = g o f.

I am not very sure how to approach this problem. I simplifed the f o g and g o f but after that I am not sure how to go from there.

Here is what I did first

f(g(x)) = g(f(x))
a(cx+d) +b = c(ax+b) +d
acx+ad+b = cax + cb +d

then after that I got stuck. I tried switching around the varibles on either side and tried to get them to equal to eachother, however that didn't work. Am I'm going in the right direction or am I totally off? Thank you for your help!!!

acx-acx = cb +d
ad+b = cb +d
ad -d = cb -b
d(a-1) = b(c-1)

that is where I got stuck

You are almost there, but may be expecting something cleverer than is
actualy the case.

First $f \circ g = g \circ f$ gives as you observe:

$acx+ad+b = cax + cb +d$,

which rearranges to:

$(ac-ca)x + ad+b-cb+d=0$,

Which holds when:

$ad+b-cb+d=0$.

Which is all there is to it; other than rearranging this last equation to as
neat a form as possible. I like:

$\frac {a-1}{b}=\frac{c-1}{d}$

RonL

3. Would it be safe to conclude that the constants a b c d may be all real numbers however a must equal to c and b must equal to d?

4. Originally Posted by hotmail590
Would it be safe to conclude that the constants a b c d may be all real numbers however a must equal to c and b must equal to d?
Yes. When a=c and b=d, the equation becomes an identity.

f(x) = ax +b
g(x) = cx +d

For f o g = g o f,
a(cx +d) +b = c(ax +b) +d
acx +ad +b = acx +bc +d
The acx cancels out,
ad +b = bc +d
ad -d = bc -b
d(a-1) = b(c-1) ---------***

So if a=c and b=d,
b(c-1) = b(c-1)
Or,
d(a-1) = d(a-1)

5. Originally Posted by hotmail590
Would it be safe to conclude that the constants a b c d may be all real numbers however a must equal to c and b must equal to d?

It is true that $a, b, c,$ and $d$ may be any real numbers which satisfy:

$<br /> \frac {a-1}{b}=\frac{c-1}{d}<br />$

So if $a=2$ and $c=4$ then any $b$ and $d$
will do as long as $d=3b$.

RonL
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