Solution
= Solution
\b[Function composition is not always commutative.] The reversed <function composition> may be undefined when the necessary <domain of a function> and <codomain> do not match. A counterexample using <functions> $\mathbb R\to\mathbb R$ avoids that issue: let $f(x)=x+1$ and $g(x)=2x$. Then
$$
(g\circ f)(x)=2x+2,\qquad(f\circ g)(x)=2x+1.
$$
Both <function compositions> exist, and their values differ at every real $x$.