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: 39

Answer

$ a.\quad g$ is one-to-one $b.\quad g^{-1}(x)=\displaystyle \frac{x+1}{3}$

Work Step by Step

$ a.\quad$ The function is linear, non-constant. Its graph is an oblique line that 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 $y.$ $y=3x-1$ 2. Interchange $x$ and $y$. (This gives the inverse function.) $x=3y-1$ 3. Solve for $y.$ $x+1=3y$ $\displaystyle \frac{x+1}{3}=y$ 4. Replace $y$ with $g^{-1}(x)$ . (This is inverse function notation.) $g^{-1}(x)=\displaystyle \frac{x+1}{3}$
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