Discrete and combinatoric mathematics (Functions)

Question

$f = ax^2 - b$ and $g = cx + d$ Where $a,b,c,d$ are all coefficients. Find $a,b,c,d$ when $f◦g = g◦f$.

Here is what I have: \begin{align*} f◦g &= a(cx + d)^2 -b = a(c^2)(x^2) + 2acdx + ad^2 - b\\ g◦f &= c(ax^2 - b) + d = acx^2 - bc + d \end{align*}

And then I get stuck... I tried manipulating variables to make them equal to each other but can't find the answer. Apparently you have to use coefficient rule, but I don't know how.

Any help is appreciated.

Answer 1

Assuming this: $(f◦g)(x)= a(cx + d)^2 -b = a(c^2)(x^2) + 2acdx + ad^2 - b$ and

$(f◦g)(x) = c(ax^2 - b) + d = acx^2 - bc + d$. As $(f◦g)(x)=(f◦g)(x)$ implies that $ac^2x^2=acx^2$, $2acdx=0x$ and $ad^2-b=-bc+d$. $ac^2x^2=acx^2$ implies that $c=1$ or $c=0$ and $a$ is any number. The second equation $2adx=0x$ implies that $a=0$ or $d=0$. The third one $ad^2-b=-bc+d$ becomes $ad^2-b=-b+d$ if $c=1$ which implies that $d=0$ and $b$ is any number. If $c=0$ then $ad^2-b=d$ implies that $b=0$, $d=1$($a=1$) or $d=0$($a$ is any number). Putting all together if $c=1$ then $a$ and $b$ are any numbers and $d=0$. If $c=0$ then $b=0$, $d=1$,$a=1$ or $b=0$, $d=0$, $a$ any number. So I think you are not lost. There are more than one answer to the problem. I just gave several of them. There are more.

Answer 2

You're on the right track. Since \begin{align} (f \circ g)(x) &= ac^2x^2 + 2acdx + (ad^2 - b) \\ (g \circ f)(x) &= acx^2 + (d - bc) \end{align} and $(f \circ g)(x) = (g \circ f)(x)$, you get a system of $3$ simultaneous equations (one for each coefficient). Hover your mouse over the boxes to reveal hints.

$$\left\{ \begin{align} ac^2 &= ac \\ 2acd &= 0 \\ ad^2 - b &= d - bc \end{align} \right.$$

Now you have to solve the equations, taking care to make all assumptions explicit. One possibility:

(1) First, suppose that $a = 0$. Then, both of the first two equations are automatically satisfied, so they give no information about the other variables. However, the third equation becomes $d = b(c-1)$.

Another possibility:

(2) Suppose that $a \ne 0$. Then, $c^2 = c$ from the first equation, so either $c = 0$ or $c = 1$.

So,

(2a) If $c = 0$, then the second equation gives nothing new, but the third equation gives $b = ad^2 - d$. (2b) Alternatively, if $c = 1$, then $d^2 = d$, so either $d = 0$ or $d = 1$.