Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises: 23

Answer

(a) $x=2$ is a critical number of $f$ and at $x=2$ there is a horizontal tangent. $x=2$ is the point where there is a local maximum of $f$. (b) $x=6$ is a critical number of $f$ and at the point where $x=6$ there is a horizontal tangent.

Work Step by Step

(a) $f'(2)=0$ and $f''(2)=-5$ $f'(2)=0$ tells us that $x=2$ is a critical number of $f$. There will be a horizontal tangent there at $x=2$. We do not know whether $f$ changes from increase to decrease or vice versa, since there is no information as to the signs of $f'$. $f''(2)=-5\lt0$. According to the Second Derivative test, this shows that $x=2$ is the point where there is a local maximum of $f$. (b) $f'(6)=0$ and $f''(6)=0$ Again, $f'(6)=0$ tells us that $x=6$ is a critical number of $f$. There will be a horizontal tangent there at $x=6$. However, $f''(6)=0$ does not give out any information since, according to the Second Derivative Test, it is inconclusive. There must be further information to know more about the graph of $f$.
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