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 302: 39

Answer

(a) $f$ is decreasing on the intervals $(-\infty,-2)\cup(0,2)$ $f$ is increasing on the intervals $(-2,0)\cup (2,\infty)$ (b) The local maximum is $f(0) = 3$ The local minima are $f(-2) = f(2) = -5$ (c) The graph is concave down on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$ The graph is concave up on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3}, \infty)$ The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{13}{9})$ and $(\frac{2\sqrt{3}}{3}, -\frac{13}{9})$ (d) We can see a sketch of the graph below.
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Work Step by Step

(a) $f(x) = \frac{1}{2}x^4-4x^2+3$ We can find the points where $f'(x) = 0$: $f'(x) = 2x^3-8x = 0$ $2x(x^2-4) = 0$ $2x(x-2)(x+2) = 0$ $x = -2,0,2$ When $x \lt -2~~$ or $0 \lt x \lt 2~~$ then $f'(x) \lt 0$ $f$ is decreasing on the intervals $(-\infty,-2)\cup(0,2)$ When $-2 \lt x \lt 0$ or $x \gt 2$ then $f'(x) \gt 0$ $f$ is increasing on the intervals $(-2,0)\cup (2,\infty)$ (b) $f(-2) = \frac{1}{2}(-2)^4-4(-2)^2+3 = -5$ $f(0) = \frac{1}{2}(0)^4-4(0)^2+3 = 3$ $f(2) = \frac{1}{2}(2)^4-4(2)^2+3 = -5$ The local maximum is $f(0) = 3$ The local minima are $f(-2) = f(2) = -5$ (c) We can find the points where $f''(x) = 0$: $f''(x) = 6x^2-8= 0$ $3x^2-4 = 0$ $x^2 = \frac{4}{3}$ $x = \pm \sqrt{\frac{4}{3}}$ $x = -\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$ The graph is concave down when $f''(x) \lt 0$ The graph is concave down on this interval: $(-\frac{2\sqrt{3}}{3},\frac{2\sqrt{3}}{3})$ The graph is concave up when $f''(x) \gt 0$ The graph is concave up on these intervals: $(-\infty,-\frac{2\sqrt{3}}{3})\cup (\frac{2\sqrt{3}}{3}, \infty)$ $f(-\frac{2\sqrt{3}}{3}) = \frac{1}{2}(-\frac{2\sqrt{3}}{3})^4-4(-\frac{2\sqrt{3}}{3})^2+3 = -\frac{13}{9}$ $f(\frac{2\sqrt{3}}{3}) = \frac{1}{2}(\frac{2\sqrt{3}}{3})^4-4(\frac{2\sqrt{3}}{3})^2+3 = -\frac{13}{9}$ The points of inflection are $(-\frac{2\sqrt{3}}{3}, -\frac{13}{9})$ and $(\frac{2\sqrt{3}}{3}, -\frac{13}{9})$ (d) We can see a sketch of the graph below.
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