Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 19

Answer

A. The domain is $(-\infty,-1)\cup (-1, \infty)$ B. The y-intercept is $0$ The x-intercept is $0$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \frac{x^3}{x^3+1} = 1$ $\lim\limits_{x \to \infty} \frac{x^3}{x^3+1} = 1$ $y = 1$ is a horizontal asymptote. $\lim\limits_{x \to -1^-} \frac{x^3}{x^3+1} = \infty$ $\lim\limits_{x \to -1^+} \frac{x^3}{x^3+1} = -\infty$ $x = -1$ is a vertical asymptote. E. The function is increasing on the intervals $(-\infty, -1)\cup (-1,0)\cup (0, \infty)$ F. There is no local maximum or local minimum. G. The graph is concave down on the intervals $(-1,0)\cup (0.794,\infty)$ The graph is concave up on the intervals $(-\infty,1)\cup (0,0.794)$ The points of inflection are $(0,0)$ and $(0.794, 0.33)$ H. We can see a sketch of the curve below.
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Work Step by Step

$y = \frac{x^3}{x^3+1}$ A. The function is defined for all real numbers except $x=-1$ The domain is $(-\infty,-1)\cup (-1, \infty)$ B. When $x = 0$, then $y = \frac{0^3}{0^3+1}= 0$ The y-intercept is $0$ When $y = 0$: $\frac{x^3}{x^3+1} = 0$ $x^3 = 0$ $x = 0$ The x-intercept is $0$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \frac{x^3}{x^3+1} = 1$ $\lim\limits_{x \to \infty} \frac{x^3}{x^3+1} = 1$ $y = 1$ is a horizontal asymptote. $\lim\limits_{x \to -1^-} \frac{x^3}{x^3+1} = \infty$ $\lim\limits_{x \to -1^+} \frac{x^3}{x^3+1} = -\infty$ $x = -1$ is a vertical asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \frac{(3x^2)(x^3+1)-(x^3)(3x^2)}{(x^3+1)^2} = \frac{3x^2}{(x^3+1)^2} = 0$ $3x^2 = 0$ $x = 0$ When $x \lt -1$ or $-1 \lt x \lt 0$ or $x \gt 0$, then $y' \gt 0$ The function is increasing on the intervals $(-\infty, -1)\cup (-1,0)\cup (0, \infty)$ F. There is no local maximum or local minimum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{(6x)(x^3+1)^2-(3x^2)(2)(x^3+1)(3x^2)}{(x^3+1)^4}$ $y'' = \frac{(6x)(x^3+1)-(3x^2)(2)(3x^2)}{(x^3+1)^3}$ $y'' = \frac{6x-12x^4}{(x^3+1)^3} = 0$ $6x-12x^4 = 0$ $-6x(2x^3-1) = 0$ $x = 0$ or $x = \sqrt[3] {\frac{1}{2}}$ $x = 0$ or $x = 0.794$ When $-1 \lt x \lt 0$ or $x \gt 0.794$ then $y'' \lt 0$ The graph is concave down on the intervals $(-1,0)\cup (0.794,\infty)$ When $x \lt -1$ or $0 \lt x \lt 0.794$, then $y'' \gt 0$ The graph is concave up on the intervals $(-\infty,1)\cup (0,0.794)$ When $x = 0$, then $y = \frac{0^3}{0^3+1}= 0$ When $x = 0.794$, then $y = \frac{0.794^3}{0.794^3+1}= \frac{1}{3}$ The points of inflection are $(0,0)$ and $(0.794, 0.33)$ H. We can see a sketch of the curve below.
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