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: 70

Answer

With $h(x)=\sqrt{x},\quad g(x)=x-1,\ \quad$and $f(x)=\sqrt[3]{x}$, $F(x)=f(g(h(x)))=(f\circ g\circ h)(x)$

Work Step by Step

If we were to calculate by hand, we would find the square root of the number $\qquad$...$ h(x)=\sqrt{x}$, then subtract 1 (from the number )$\qquad$ .... $g(x)=x-1$ and then find the cube root $\qquad$ .... $f(x)=\sqrt[3]{x}$ In this way, $F(x)=f(g(h(x)))=f\circ g\circ h(x)$
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