Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.7 - Combining Functions - 2.7 Exercises - Page 217: 67

Answer

With $g(x)=1-x^{3}$, and $f(x)=|x|$, $H(x)=f[g(x)]=f\circ g(x)$

Work Step by Step

If we were to calculate by hand, we would subtract the cube of the number from 1$\qquad$...$ g(x)=1-x^{3}$, then find its absolute value$\qquad$ .... $f(x)=|x|$ In this way, $H(x)=f(g(x))=f\circ g(x)$
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