College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 6 - Section 6.4 - Logarithmic Functions - 6.4 Assess Your Understanding - Page 449: 48

Answer

$(-\infty,0)\cup(1,\infty)$ or $\{x|x \lt 0$ or $x \gt 1\}$

Work Step by Step

First, we see that if we want $\displaystyle \frac{x}{x-1}$ to be defined, we must have $x\neq 1\quad $ Next, the argument of a logarithmic function must be greater than zero. So, for h to be defined, we must have $\displaystyle \frac{x}{x-1} \gt 0 \quad $ ... $x-1$ changes signs at $1$, x changes signs at 0 $Q(x)=\displaystyle \frac{x}{x-1}$ is zero or undefined for $x=0,1$. We test intervals into which these two numbers split the number line. $\left[\begin{array}{llll} \text{interval} & (-\infty,0) & (0,1) & (1,\infty)\\ \text{test point } t & -1 & 0.5 & 2\\ Q(t) & \frac{-1}{-1-1} & \frac{0.5}{0.5-1} & \frac{2}{2-1}\\ \text{sign of }Q(t) & + & - & + \end{array}\right]$ Domain: $(-\infty,0)\cup(1,\infty)$ or $\{x|x \lt 0$ or $x \gt 1\}$
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