## Precalculus (6th Edition) Blitzer

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.