## Intermediate Algebra (6th Edition)

$(f\circ g)(x)=(x+3)^2$ $(g \circ f)(x) = x^2+3$
RECALL: (1) $(f \circ g)(x) = f[g(x)]$ (2) $(g \circ f)(x) = g[f(x)]$ Thus, $(f\circ g)(x)= f[(g(x)]$ $\\(f\circ g)(x)=(x+3)^2$ (Substitute $x+3$ to $x$ in $f(x)$) and $(g \circ f)(x) = g[f(x)]$ $(g \circ f)(x) = x^2+3$ (Substitute $x^2$ to $x$ in $g(x)$)