Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 70

Answer

$(-\infty, 6)$

Work Step by Step

(i) The denominator cannot be equal to zero thus, $\sqrt{6-x}\ne 0 \\6-x \ne 0^2 \\6-x \ne 0 \\6 \ne x$ (ii) The radicand (number inside the square root symbol) cannot be negative. Thus, $6-x \ge 0 \\-x \ge -6 \\-x(-1) \le -6(-1) \\x \le 6$ This means that the value of $x$ has to be less than 6. Therefore, the domain of the function is: $(-\infty, 6)$
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