Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 1 - Section 1.1 - Functions from the Numerical, Algebraic, and Graphical Viewpoints - Exercises - Page 52: 18e

Answer

$h(x^2+1)=\frac{1}{x^2+5}$

Work Step by Step

$h(r)=\frac{1}{r+4}$ Substitute $r$ with $x^2+1$ $h(x^2+1)=\frac{1}{x^2+1+4}=\frac{1}{x^2+5}$
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