Suppose that $$f(x) = ax^2 +b$$ is a quadratic function, where $ (a, b) \in \mathbb R^2$ and $a \neq 0. $ If $$g(x) = cx^2 +d,$$ where $(c, d) \in \mathbb R^2$ and $c \neq 0,$ is another quadratic function that commutes with $f(x)$
-- i.e., $(f\circ g)(x) = (g \circ f)(x)$ for all $x \in \mathbb R$) -- then $$f(x) = g(x)$$ for all $x \in \mathbb R.$ I'm not sure how to prove this. I don't even know where to start; I am very stuck. If someone could please help, I really need it. I need to prove that $f(x)=g(x).$
$f \circ g = a (g(x))^2 + b = a (cx^2 + d)^2 + b = ac^2x^4 + 2acdx^2 + ad^2 + b$
$g \circ f = c (f(x))^2 + d = c (ax^2 + b)^2 + d = ca^2x^4 + 2cabx^2 + cb^2 + d$
Therefore, by identifying the coefficient, $ca^2 = ac^2$ and $2acd = 2 abc$ and $ad^2 + b = cb^2 + d$.