Answer
The statement is false. The correct statement is, “There can be two functions f and g, where $f\ne \text{g}$ , for which $\left( f\circ g \right)\left( x \right)=\left( g\circ f \right)\left( x \right)$.”
Work Step by Step
There exist two functions, such that $f\left( g\left( x \right) \right)=g\left( f\left( x \right) \right)$.
Take $f\left( x \right)=x,\text{ g}\left( x \right)=\frac{1}{x}$
Find:
$\begin{align}
& f\left( g\left( x \right) \right)=g\left( x \right) \\
& =\frac{1}{x}
\end{align}$
$\begin{align}
& g\left( f\left( x \right) \right)=\frac{1}{f\left( x \right)} \\
& =\frac{1}{x}
\end{align}$
Hence, $f\left( g\left( x \right) \right)=g\left( f\left( x \right) \right)$
The given statement “There can never be two functions f and g, where $f\ne \text{g}$ ,for which $f\left( g\left( x \right) \right)=g\left( f\left( x \right) \right)$ ” is false.
