Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 7 - Section 7.1 - Radical Expressions and Functions - Exercise Set - Page 511: 25

Answer

$h(5)=4$ $h(3)=2$ $h(0)=1$ $h(-5)=6$.

Work Step by Step

The given function is $h(x)=\sqrt{(x-1)^2}$ Plug $x=5$ into the function. $h(5)=\sqrt{(5-1)^2}$ Simplify. $h(5)=\sqrt{(4)^2}$ $h(5)=\sqrt{4^2}$ $h(5)=4$ Plug $x=3$ into the function. $h(3)=\sqrt{(3-1)^2}$ Simplify. $h(3)=\sqrt{(2)^2}$ $h(3)=\sqrt{2^2}$ $h(3)=2$ Plug $x=0$ into the function. $h(0)=\sqrt{(0-1)^2}$ Simplify. $h(0)=\sqrt{(-1)^2}$ $h(0)=\sqrt{1^2}$ $h(0)=1$ Plug $x=-5$ into the function. $h(-5)=\sqrt{(-5-1)^2}$ Simplify. $h(-5)=\sqrt{(-6)^2}$ $h(-5)=\sqrt{6^2}$ $h(-5)=6$.
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