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 - Page 304: 89

Answer

The point $(0,0)$ is an inflection point, but $g''(0)$ does not exist.

Work Step by Step

$g(x) = x~\vert x \vert$ $g(x) = x^2~~~~~$ for $x \geq 0$ $g(x) = -x^2~~~~~$ for $x \lt 0$ $g'(x) = 2x~~~~~$ for $x \geq 0$ $g'(x) = -2x~~~~~$ for $x \lt 0$ $g''(x) = 2~~~~~$ for $x \gt 0$ $g''(x) = -2~~~~~$ for $x \lt 0$ However, $g''(0)$ does not exist since $\lim\limits_{h \to 0^-} g''(0+h) \neq \lim\limits_{h \to 0^+} g''(0+h)$ The graph is concave downward when $x \lt 0$ The graph is concave upward when $x \gt 0$ The graph changes from concave downward to concave upward at the point $(0,0)$, The point $(0,0)$ is an inflection point, but $g''(0)$ does not exist.
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