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I want to ask a follow up question to this one: 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 $f ◦ g = g ◦$

I thought the question was trying to establish an identity function:

I solve that $f \circ g= g \circ f$

$acx+ad+b=acx+bc+d$ factored to

$d(a-1)=b(c-1)$

So this mean that the following conditions are necessary and sufficient: $a \neq 1 \land b \neq 0 \land c \neq 1 \land d \neq 0$

But if I let $f^{-1}(x) = \frac{x}{a} - \frac {b}{a}$ and $g^{-1}(x) = \frac{x}{c} - \frac{d}{c}$ are the conditions $a \neq 0 \land c \neq 0 $ also necessary and sufficient?

Elliott de Launay
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  • The expression $f \circ g = g \circ f$ does not involve the function inverses, so there is no need to divide by $a$ or $c$. – angryavian Feb 13 '19 at 01:40
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    I think your conclusion [before your last sentence] is also not right; just stay with the condition given in the answer to your other question. – angryavian Feb 13 '19 at 01:42
  • @angryavian I was thinking if I wanted to solve for any of constants for example $d$ then $(a-1)$ could not be $0$ and vice versa. – Elliott de Launay Feb 13 '19 at 01:46
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    The question is not asking you to solve for the constants; there will be many different possible solutions. They are just asking you to find a condition on the constants. – angryavian Feb 13 '19 at 01:48

1 Answers1

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$(f\circ g)(x) = a(cx + d) + b = ac x +ad + b\\ (g\circ f)(x) = c(ax + b) + d = ac x +bc + d$

$(f\circ g)(x) = (g\circ f)(x) \implies bc + d = ad +b$ or $d(a-1) = b(c-1)$

$f$ or $g$ is the identity function. i.e. $a= 1$ and $b = 0$ and $f(x) = x$

or

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

Doug M
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