College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 192: 22


$h(2)=\frac{8}{5}$ $h(-2)=\frac{8}{5}$ $h(a)=\frac{a^{2}+4}{5}$ $h(-x)=\frac{x^{2}+4}{5}$ $h(a-2)=\frac{a^{2}-4a+8}{5}$ $h(\sqrt{x})=\frac{x+4}{5}$

Work Step by Step

We are given: $h(x)= \frac{x^{2}+4}{5}$ We evaluate: $h(2)=\frac{2^{2}+4}{5}=\frac{4+4}{5}=\frac{8}{5}$ $h(-2)=\frac{(-2)^{2}+4}{5}=\frac{8}{5}$ $h(a)=\frac{a^{2}+4}{5}$ $h(-x)=\frac{(-x)^{2}+4}{5}=\frac{x^{2}+4}{5}$ $h(a-2)= \frac{(a-2)^{2}+4}{5}=\frac{a^2-4a+4+4}{5}=\frac{a^{2}-4a+8}{5}$ $h(\sqrt{x})=\frac{(\sqrt{x})^{2}+4}{5}=\frac{x+4}{5}$
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