Answer
false
Work Step by Step
RECALL:
$(f \circ g)=f(g(x))$
This means that $g(x)$ will replce every $x$ in $f(x)$.
To illustrate how composition works, have a look at the example below:
Let
$f(x) = x+2$
$g(x) = 2x$
$(f \circ g)(x) = f(g(x)) = 2x + 2$
Note that
$f(x) \cdot g(x) = (x+2) \cdot 2x = 2x^2 + 4x$
Thus, $f(g(x)) \ne f(x) \cdot g(x)$.
The given statement is false.
