Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review: 21b

Answer

$(f\circ g)(x)=(x+3)^2$ $(g \circ f)(x) = x^2+3$

Work Step by Step

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 $g(x)$)
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