Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Review - Exercises - Page 690: 6

Answer

The graph for $y=\pm x \sqrt {x+1}$ is as shown. As $t$ increases, the curve is traced from the bottom right, to the top left, then bottom left, and finally top right.

Work Step by Step

Finding $f(t)$: It is a vertical parabola with zeroes at $t=\pm 1$ thus, $f(t)=k(t-1)(t+1)=k(t^2-1)$ from the graph, we have $f(0)=-1$ hence $k=1$ $f(t)=(t^2-1)=x$ ... (1) we can also write $t=\pm \sqrt {x+1}$ ... (2) Finding $g(t)$ It is a cubical polynomial with zeroes at $t=0,\pm 1$ thus, $g(t)=k(t-1)(t+1)=k(t^2-1)$ from the graph, we have $g(0.5)=-0.4$ hence $k=1$ $g(t)=(t^2-1)=y$ we can also write $y=\pm x \sqrt {x+1}$ The graph for $y=\pm x \sqrt {x+1}$ is as shown.
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