Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.1 - Four Ways to Represent a Function - 1.1 Exercises - Page 22: 56

Answer

$$ f(x)= \left\{ \begin{array}{3} -\frac{3}{2} x - 3 & \quad -4 \le x \le -2 \\ \sqrt{4 - x^2} & \quad -2 \lt x \lt 2 \\ \frac{3}{2} x- 3 & \quad 2 \le x \le 4 \\ \end{array} \right. $$

Work Step by Step

One will notice that the graph appears to be an absolute value function split into 2 by a semi circle. The absolute value function has a slope of $-\frac{3}{2} $ and $+\frac{3}{2} $ and a y-intercept of $-3$. So the equation of this line is $y = \pm \frac{3}{2} x \ -3$. The semi circle has a radius of $2$ and a center at $(0, \ 0)$. Therefore, its equation would be $y= \sqrt{4 - x^2}$. This equation's domain is $(-2, 2)$.
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