Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises: 22

Answer

Evaluate $h(x) = \frac{x^{2}+4}{5}$: $h(2) = 1.6 $ $h(-2) = 1.6$ $h(a) = \frac{a^{2}+4}{5}$ $h(-x) = \frac{x^{2}+4}{5}$ $h(a-2) = \frac{a^{2}-4a+ 8}{5}$ $h(\sqrt x) = \frac{x+4}{5}$

Work Step by Step

x = 2 $h(2) = \frac{2^{2}+4}{5}$ ... square 2 $= \frac{4+4}{5}$ ... add numerator $= \frac{8}{5}$ ... divide $=1.6$ ---------------------- x = -2 $h(-2) = \frac{(-2)^{2}+4}{5}$ ... square -2 $= \frac{4+4}{5}$ ... add numerator $= \frac{8}{5}$ ... divide $=1.6$ ---------------------- x = a $h(a) = \frac{a^{2}+4}{5}$ ---------------------- x = -x $h(-x) = \frac{(-x)^{2}+4}{5}$ ... square -x $= \frac{x^{2}+4}{5}$ ---------------------- x = a-2 $h(a-2) = \frac{(a-2)^{2}+4}{5}$ ... square (a-2) $= \frac{(a^{2}-4a+4)+4}{5}$ ... add numerator $= \frac{a^{2}-4a+ 8}{5}$ ---------------------- x = $\sqrt x$ $h(\sqrt x) = \frac{(\sqrt x)^{2}+4}{5}$ ... square $\sqrt x$ $= \frac{x+4}{5}$ ----------------------
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