Question

The composition of functions is commutative.

Answer 1

Define functions and verify to get the desired result.

Let $f\left(x\right)=x^2$ and $g\left(x\right)=x+1$
$\left(fog\right)\left(x\right)=f\left(g\right(x\left)\right)$
$=f(x+1)$
$=(x+1{)}^2$
$=x^2+2x+1\dots \left(1\right)$
$\left(gof\right)\left(x\right)=g\left(f\right(x\left)\right)$
$=g\left(x^2\right)$
$=x^2+1\dots \left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right)$, we get
$\left(fog\right)\left(x\right)\ne \left(gof\right)\left(x\right)$
Therefore, the composition of function $f$ and $g$ is not commutative.

Hence, the given statement is false.