Intermediate Algebra (6th Edition)

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

Chapter 11 - Cumulative Review - Page 670: 24a



Work Step by Step

With $f(x)=x^2-2$ and $g(x)=x+1$, then, \begin{array}{l} (f\circ g)(2) \\= f(g(2)) \\= f(2+1) \text{...substitute $x=2$ in $g$} \\= f(3) \\= 3^2-2 \text{...substitute $x=3$ in $f$} \\= 7 .\end{array}
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