Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - Review - Exercises: 5

Answer

Domain: $(-\infty, 1/3)\cup(1/3, \infty)$ Range: $(-\infty, 0)\cup(0, \infty)$

Work Step by Step

Let $f(x) = \frac{2}{3x-1}$ This function is defined for all values of $x$, except when $3x-1 = 0 \implies 3x = 1 \implies x = \frac{1}{3}$ Since this value cancels the denominator, therefore the function is not defined for $x = \frac{1}{3}$ Thus, the domain of $f = (-\infty, 1/3)\cup(1/3, \infty)$ Given $f(x) = y$ $y = \frac{2}{3x-1} \implies y(3x-1) = 2 \implies 3xy-y = 2 \implies 3xy = y + 2 \\[0em] \implies x = \frac{y+2}{3y}$ This expression is defined for all values of $y$, except when $3y = 0 \implies y = 0$, for the same reasoning as above. Therefore, the range of $f = (-\infty, 0)\cup(0, \infty)$ See the attached figure.
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