Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 29

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.

$y = x-3x^{1/3}$ 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.