#### Answer

false

#### Work Step by Step

RECALL:
(1) $(f \circ g)-f(g(x))$
(2) $(g \circ f)(x) = g(f(x))$
It often happens that $(f \circ g)(x) \ne (g \circ f)(x)$.
To illustrate, have a look at the example below:
Let
$f(x) = x+2$
$g(x) = 2x$
$(f \circ g)(x) = f(g(x)) = 2x + 2$
$(g \circ f)(x) = g(f(x)) = 2(x+2) =2x+4$
Thus, the given statement is false.