Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Review - Exercises - Page 322: 101

Answer

[-3, 8/3]

Work Step by Step

The question asks for the domain of the function given in the problem Given $f(x) = \sqrt {24 - x - 3x^2}$ The domain is such that $24 - x - 3x^2$ is greater than or equal to 0. $24 - x - 3x^2 \geq 0$ $3x^2 + x - 24 \leq 0$ $(3x - 8) (x + 3) \leq 0$ $x = 8/3, -3$ (-∞, -3] $(-)(-) = (+)$ [-3, 8/3] $(+)(-) = (-)$ [8/3, ∞) $(+)(+) = (+)$ Thus the Domain is [-3, 8/3]
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