Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Practice Exercises - Page 179: 91

Answer

See graph and explanations.

Work Step by Step

Step 1. Based on the figure and conditions given in the Exercise, the first line segment starts at point $(-1,2)$ with a slope of $-2$ (read from the step function); thus its line equation is $y-2=-2(x+1)$ or $y=-2x$. This also gives its other endpoint at $(0,0)$ Step 2. The second line segment starts at $(0,0)$ and has a slope of $0$; thus its equation is $y-0=0(x-0)$ or $y=0$, which is the x-axis. The other endpoint for this segment is $(1,0)$ Step 3. The third line segment starts at $(1,0)$ and has a slope of $1$. Thus its equation is $y=x-1$, and the other endpoint for this segment is $(4,3)$ Step 4. The last line segment starts at $(4,3)$ and has a slope of $-1$. Thus its equation is $y-3=-(x-4)$ or $y=-x+7$, and the other endpoint for this segment is $(6,1)$ Step 5. We can graph the above results as shown in the figure.
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