Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.1 Properties of Functions - 2.1 Exercises - Page 54: 31


The domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .

Work Step by Step

$$ f(x)=\frac{1}{\sqrt {3x^{2}+2x-1}}=\frac{1}{\sqrt {(3x-1)(x+1)}} $$ the values $x$ for $f(x)$ is defined when: $$ (3x-1)(x+1) \gt 0 $$ since the radical cannot be negative and the denominator of the function cannot be zero. So we can solve: $$ (3x-1)(x+1) = 0 $$ $ \Rightarrow $ $$ x_{1}=\frac{1}{3},\:x_{2}=-1 $$ this numbers $-1$ and $ \frac{1}{3}$ divide the number line into 3 intervals, $ (-\infty,-1), (-1 ,\frac{1}{3}) $ and $(\frac{1}{3},\infty)$. we observe that only values in the intervals $(-\infty,-1) $ and $(\frac{1}{3},\infty)$ satisfy the inequality. So the domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
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