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: 58

Answer

$ a.\quad f$ is one-to-one $b.\quad f^{-1}(x)=x^{2}+1,\qquad x\geq 0$

Work Step by Step

$ a.\quad$ The graph of $f$ passes the horizontal line test (see below.) (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. Note that $x\geq 1$ and $f(x)\geq 0.$. The domain of $f^{-1}(x)$ is nonnegative numbers x. $ b.\quad$ To find a formula for the inverse, 1. Replace $g(x)$ with $y.$ $y=\sqrt{x-1}, \qquad x\geq 1, y\geq 0$ 2. Interchange $x$ and $y$. (This gives the inverse function.) $x=\sqrt{y-1}, \qquad y\geq 1, x\geq 0$ 3. Solve for $y.$ ... square both sides, $ x^{2}=y-1,\qquad y\geq 1, x\geq 0\qquad$ ... add 1 $x^{2}+1=y,\qquad y\geq 1, x\geq 0$ 4. Replace $y$ with $f^{-1}(x)$ . (This is inverse function notation.) $f^{-1}(x)=x^{2}+1,\qquad x\geq 0$
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