Intermediate Algebra (6th Edition)

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

Chapter 11 - Cumulative Review - Page 670: 21a


$(f \circ g)(2) = 25$ $(g \circ f)(2) = 7$

Work Step by Step

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