Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.1 Functions - Exercises Set 0.1: 17

Answer

Yes. $$y= \begin{cases} \sqrt{25-x^2} & \text{ if } -5 \le x \le 0 \\ - \sqrt{25-x^2} & \text{ if } 0 < x \le 5 \end{cases}$$

Work Step by Step

Yes, it is a function since there exists no real number for which there are two distinct values of $y$. This graph consists of two sections of the circle $x^2+y^2=25$: First, the left half of the upper semicircle $y= \sqrt{25-x^2}$, including the points $(0,5)$ and $(-5,0)$, that is,$$y=\sqrt{25-x^2} \qquad -5 \le x \le 0.$$Second, the right half of the lower semicircle $x^2+y^2=25$, including the point $(5,0)$ but not the point $(0,-5)$, that is,$$y=- \sqrt{25-x^2} \qquad 0 \le x \le 5.$$ Thus, the function representing this graph is$$y= \begin{cases} \sqrt{25-x^2} & \text{ if } -5 \le x \le 0 \\ - \sqrt{25-x^2} & \text{ if } 0 < x \le 5 \end{cases}.$$
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