Answer
$0\leq p\leq4$
Work Step by Step
$\because$ the square root of non-negative numbers is real,
i. $p\geq0$
ii. $2-\sqrt p\geq0, \sqrt p\leq2, p\leq2^{2}=4$
$\therefore 0\leq p\leq4$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises - Page 21: 37
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