## Precalculus (6th Edition) Blitzer

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.