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: 40

Answer

(a) $g$ is decreasing on the interval $(-\infty,-6)$ $g$ is increasing on the interval $(-6,\infty)$ (b) The local minimum is $g(-6) = -232$ There is no local maximum. (c) The graph is concave down on this interval: $(-4,0)$ The graph is concave up on these intervals: $(-\infty,-4)\cup (0, \infty)$ The points of inflection are $(-4, -56)$ and $(0, 200)$ (d) We can see a sketch of the graph below.
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Work Step by Step

(a) $g(x) = 200+8x^3+x^4$ We can find the points where $g'(x) = 0$: $g'(x) = 24x^2+4x^3 = 0$ $4x^2(6+x) = 0$ $x = 0, -6$ When $x \lt -6~~$ then $g'(x) \lt 0$ $g$ is decreasing on the interval $(-\infty,-6)$ When $-6 \lt x~~$ then $g'(x) \gt 0$ $g$ is increasing on the interval $(-6,\infty)$ (b) $g(-6) = 200+8(-6)^3+(-6)^4 = -232$ The local minimum is $g(-6) = -232$ There is no local maximum. (c) We can find the points where $g''(x) = 0$: $g''(x) = 48x+12x^2= 0$ $12x(4+x)= 0$ $x = -4, 0$ The graph is concave down when $g''(x) \lt 0$ The graph is concave down on this interval: $(-4,0)$ The graph is concave up when $g''(x) \gt 0$ The graph is concave up on these intervals: $(-\infty,-4)\cup (0, \infty)$ $g(-4) = 200+8(-4)^3+(-4)^4 = -56$ $g(0) = 200+8(0)^3+(0)^4 = 200$ The points of inflection are $(-4, -56)$ and $(0, 200)$ (d) We can see a sketch of the graph below.
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