## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises - Page 304: 88

#### Answer

$(0,0)$ is not an inflection point of the graph $f$.

#### Work Step by Step

$f(x) = x^4$ $f'(x) = 4x^3$ $f''(x) = 12x^2$ $f''(x) = 0$ when $x = 0$ $f(0) = 0$, so the point $(0,0)$ is a point on the graph of $f(x)$ However: $f''(x) \gt 0$ when $x \lt 0$ $f''(x) \gt 0$ when $x \gt 0$ Therefore, $f(x)$ is concave upward. At the point $(0,0)$, the graph does not change from concave upward to concave downward (or vice versa). Therefore, $(0,0)$ is not an inflection point of the graph $f$.

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