## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 0 - Before Calculus - 0.2 New Functions From Old - Exercises Set 0.2 - Page 25: 29

#### Answer

$a. f(g(2))= 3$ $b. g(f(4))= 9$ $c. f(f(16))= 2$ $d. g(g(0))= 2$ $e. f(2+h)=\sqrt {(2+h)}$ $f. g(3+h)= (3+h)^{3}+1$

#### Work Step by Step

$a. f(g(2)) = f(2^{3}+1) =\sqrt (2^{3}+1) =\sqrt 9 =3$ $b. g(f(4)) =g(\sqrt 4) =g(2) =2^{3}+1 =9$ $c. f(f(16)) =f(\sqrt 16) =f(4) =\sqrt 4 =2$ $d. g(g(0)) =g(1) =2$ $e. f(2+h) =\sqrt {(2+h)}$ $f. g(3+h) =(3+h)^{3}+1$

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