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: 29

Answer

$(-\infty,-1] \cup [5,\infty)$

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 radicand must not be negative, $x^{2}-4x-5 \geq 0$ Factoring the LHS, we find two factors of -5 whose sum is -4. (These are $-5$ and $+1)$ $(x+1)(x-5) \geq$ 0 The zeros of the LHS, $x=-1$ and $x=5$ divide the real number line into intervals. Interval: $(-\infty,-1]$, test point x=$-2$, $(-2+1)(-2-5)=-1(-7)=+7$ (positive) Interval: $(-1, 5)$, test point x=$0$, $(0+1)(0-5)=1(-5)=-5$ (negative) Interval $[5,\infty)$, test point x=$10$, $(10+1)(10-5)=11(5)=55$ (positive) So, the domain is $(-\infty,-1] \cup [5,\infty)$
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