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: 29

Answer

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $0$ The x-intercepts are $-5.2, 0, 5.2$ C. The function is an odd function. D. $\lim\limits_{x \to -\infty} x-3x^{1/3} = -\infty$ $\lim\limits_{x \to \infty} x-3x^{1/3} = \infty$ No asymptotes. E. The function is decreasing on the interval $(-1,1)$ The function is increasing on the intervals $(-\infty, -1)\cup (1, \infty)$ F. The local maximum is $(-1, 2)$ The local minimum is $(1, -2)$ G. The graph is concave down on the interval $(-\infty,0)$ The graph is concave up on the interval $(0, \infty)$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.
1559051779

Work Step by Step

$y = x-3x^{1/3}$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x=0$, then $y = 0-3(0)^{1/3} = 0$ The y-intercept is $0$ When $y = 0$: $x-3x^{1/3} = 0$ $x=3x^{1/3}$ $x^3=27x$ $x^3-27x = 0$ $x(x^2-27) = 0$ $x = -\sqrt{27}, 0, \sqrt{27}$ $x = -5.2, 0, 5.2$ The x-intercepts are $-5.2, 0, 5.2$ C. $(-x)-3(-x)^{1/3}= -(x-3x^{1/3})$ The function is an odd function. D. $\lim\limits_{x \to -\infty} x-3x^{1/3} = -\infty$ $\lim\limits_{x \to \infty} x-3x^{1/3} = \infty$ There are no horizontal asymptotes. E. We can find values of $x$ such that $y' = 0$: $y' =1-x^{-2/3} = 1-\frac{1}{x^{2/3}} = 0$ $x^{2/3} = 1$ $x = -1, 1$ Note that $y'$ is undefined when $x=0$ When $-1 \lt x \lt 0$ or $0 \lt x \lt 1$, then $y' \lt 0$ The function is decreasing on the interval $(-1,1)$ When $x \lt -1$ or $x \gt 1$, then $y' \gt 0$ The function is increasing on the intervals $(-\infty, -1)\cup (1, \infty)$ F. When $x=-1$, then $y = (-1)-3(-1)^{1/3} = 2$ The local maximum is $(-1, 2)$ When $x=1$, then $y = (1)-3(1)^{1/3} = -2$ The local minimum is $(1, -2)$ G. We can find the values of $x$ such that $y'' = 0$: $y'' =\frac{2}{3}\cdot \frac{1}{x^{5/3}} = \frac{2}{3~x^{5/3}} = 0$ There are no values $x$ such that $y'' = 0$ Note that $y''$ is undefined when $x=0$ When $x \lt 0$, then $y'' \lt 0$ The graph is concave down on the interval $(-\infty,0)$ When $x \gt 0$, then $y'' \gt 0$ The graph is concave up on the interval $(0, \infty)$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.
Small 1559051779
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.