Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.3 - New Functions from Old Functions - 1.3 Exercises - Page 45: 63

Answer

(a) $f(x)=x^2+6$ (b) $g(x)=x^2+x-1$

Work Step by Step

(a) To find $f$ such that $f \circ g=h$, we have$$f(g(x))=h(x) \quad \Rightarrow \quad f(2x+1)=4x^2+4x+7$$ $$\Rightarrow \quad f(2x+1)=(2x+1)^2+6$$ $$\Rightarrow \quad f(x)=x^2+6.$$ (b) To find $g$ such that $f \circ g=h$, we have$$f(g(x))=h(x) \quad \Rightarrow \quad 3g(x)+5=3x^2+3x+2$$ $$\Rightarrow \quad 3g(x)+5=3(x^2+x-1)+5$$ $$\Rightarrow \quad g(x)=x^2+x-1$$
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