Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 65: 50

Answer

(a) The domain of $f(x)$ is $(1, \infty)$ The range of $f(x)$ is $(-\infty, \infty)$ (b) The x-intercept is $~~e+1$ (c) We can see a sketch of the graph of $f$ below.

Work Step by Step

(a) The domain of $ln(x)$ is $x \gt 0$ We can find the domain of $f(x) = ln(x-1)-1$: $x-1 \gt 0$ $x \gt 1$ The domain of $f(x)$ is $(1, \infty)$ The range of $ln(x)$ is $(-\infty, \infty)$ Therefore the range of $f(x)$ is $(-\infty, \infty)$ (b) We can find the x-intercept: $f(x) = ln(x-1)-1 = 0$ $ln(x-1)=1$ $x-1=e^1$ $x=e+1$ The x-intercept is $~~e+1$ (c) We can see a sketch of the graph of $f$ below.
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