Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set: 27

Answer

$[2, 5) \cup (5, +\infty)$

Work Step by Step

(i) The radicand (expression inside a radical sign) of a square root cannot be negative as its root is an imaginary number. This means that the radicand in the numerator, which is $x-2$, must be greater than or equal to 0. Thus, $x$ can be any real number greater than or equal to $2$. (ii) The denominator cannot be zero as division of zero leads to an undefined expression. This means that the value of $x$ cannot be $5$. From (i) and (ii) above, the restrictions to the value of $x$ are: (i) $x \ge 2$; and (ii) $x \ne 5$ Both of the restrictions above must be satisfied. Thus, the value of $x$ must be greater than or equal to $2$ but not including $5$. Therefore, the domain of the given function is: $[2, 5) \cup (5, +\infty)$
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