## Calculus: Early Transcendentals 8th Edition

$0\leq p\leq4$
$\because$ the square root of non-negative numbers is real, i. $p\geq0$ ii. $2-\sqrt p\geq0, \sqrt p\leq2, p\leq2^{2}=4$ $\therefore 0\leq p\leq4$