Calculus 8th Edition

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

Chapter 2 - Derivatives - 2.2 The Derivative as a Function - 2.2 Exercises - Page 126: 3

Answer

See the graphs
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Work Step by Step

Analyzing the graph: Step 1: Let take a look in quadrant 2 and let us choose arbitrary x and y points Step 2: $x_{2}>x_{1}$ and $y_{2}0$ and $y_{2}-y_{1}<0$ Step4: as show any computed difference value of y will end in quadrant 3 in the black for the same domain values. Step 5: as we get $y_{2}=y_{1}$, then slope is zero. That is when we reach the x axis and that is when the red line starts. Step 6: The red section labeled on the graph says that while it is true that $y_{2}-y_{1}<0$ before the slope is zero,starting from that point on we have: $x_{2}-x_{1}>0$ and $y_{2}-y_{1}>0$ Which means as shown any computed difference value of y will end in quadrant 2 in the for the same domain values until $y_{2}-y_{1}=0$, slope is zero. Step7: In quadrant 1: $x_{2}-x_{1}>0$ and $y_{2}-y_{1}>0$. As the blue label states, any compute difference y values will end in quadrant 1 until $y_{2}-y_{1}=0$, slope is zero. Step8 Quadrant 4: $x_{2}-x_{1}>0$ and $y_{2}-y_{1}<0$ As shown in green label. Following the same steps #1-8, it slould be easy to show the derivatives of the graphs for: B, C, and D.
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