Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 12 - Exponential Functions and Logarithmic Functions - 12.1 Composite Functions and Inverse Functions - 12.1 Exercise Set - Page 787: 55

Answer

$ a.\quad g$ is one-to-one $b.\quad g^{-1}(x)=\sqrt[3]{x}+2$

Work Step by Step

$ a.\quad$ The graph of $g$ is obtained by shifting $y=x^{3}$ by 2 units to the right. (similar to example 8, where it was raised 2 units). Its graph passes the horizontal line test (It is impossible to draw a horizontal line that intersects a function's graph more than once.) It is one-to-one and has an inverse. $ b.\quad$ To find a formula for the inverse, 1. Replace $g(x)$ with $y.$ $y=(x-2)^{3}$ 2. Interchange $x$ and $y$. (This gives the inverse function.) $x=(y-2)^{3}$ 3. Solve for $y.$ ... take the cube root, $\sqrt[3]{x}=y-2\qquad$... add $2$, $\sqrt[3]{x}+2=y$ 4. Replace $y$ with $g^{-1}(x)$ . (This is inverse function notation.) $g^{-1}(x)=\sqrt[3]{x}+2$
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