Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.4 Logarithmic Functions - 4.4 Assess Your Understanding - Page 321: 74

Answer

(a) $(-\infty,0)$. (b) See graph. (c) range $(-\infty,\infty)$, V.A. $x=0$. (d) $ f^{-1}(x)=-e^{-x} $ (e) domain $(-\infty,\infty)$, range $(-\infty,0)$. (f) See graph.

Work Step by Step

(a) Given $f(x)=-ln(-x)$, we can find the domain $-x\gt0$ or $(-\infty,0)$. (b) See graph. (c) From the graph, we can determine the range $(-\infty,\infty)$, asymptote(s) V.A. $x=0$. (d) $f(x)=-ln(-x) \Longrightarrow y=-ln(-x) \Longrightarrow x=-ln(-y) \Longrightarrow y=-e^{-x} \Longrightarrow f^{-1}(x)=-e^{-x} $ (e) For $ f^{-1}(x)$, we can find the domain $(-\infty,\infty)$, range $(-\infty,0)$. (f) See graph.
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