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: 27

Answer

$(-\infty,-4)\cup(4,\infty)$

Work Step by Step

The domain is the set of all possible values of x for which the value f(x) exists (is defined). For f(x) to be defined, the following conditions must be met: 1. The radicand $\displaystyle \frac{2}{x^{2}-16}$ must not be negative (because of the square root) Since the numerator is positive, the denominator must not be negative: So, $\qquad x^{2}-16 \geq 0.$ 2. The denominator must not be zero $x^{2}-16\neq 0$ Combined, these two conditions produce $x^{2}-16 > 0$ $x^{2} > 16$ This is true for $x < -4$ and for $x > 4.$ In interval notation, the domain is $(-\infty,-4)\cup(4,\infty)$
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