Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 22: 56

Answer

$f(x)=$\begin{cases} \sqrt{1-x^2} & |x|\le2\\ |\frac{2x}{3}|-3 & |x|>2 \\ \end{cases}

Work Step by Step

Every image of x less than 2 or greater than 2 is equal to the graph of the function $f(x) = |\frac{2x}{3}|-3$, which is the absolute value of the line $y = \frac{2x}{3}$ displaced three units downwards (subtract 3), and every image of x in the interval [-2,2] is equal to the graph of the function $g(x) = \sqrt{1-x^2}$, which is the upper half of the circle of radius 2 centered at the origin (given by $x^2+y^2=4$). See the attached figure.
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