Intermediate Algebra (6th Edition)

$(f \circ g)(2) = 25$ $(g \circ f)(2) = 7$
RECALL: (1) $(f \circ g)(x) = f[g(x)]$ (2) $(g \circ f)(x) = g[f(x)]$ Thus, $(f\circ g)(x)= f[(g(x)] \\(f\circ g)(x)=(x+3)^2$ and $(g \circ f)(x) = g[f(x)]$ $(g \circ f)(x) = x^2+3$ Substitute $2$ into $x$ to obtain: $(f\circ g)(x)=(x+3)^2 \\(f\circ g)(2)=(2+3)^2 = 5^2 = 25$ and $(g \circ f)(x) = x^2+3 \\$(g \circ f)(x) = 2^2+3=4+3=7\$