Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 22

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $0$ The x-intercepts are $0$ and $4$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} (x-4)~\sqrt[3] x = \infty$ $\lim\limits_{x \to \infty} (x-4)~\sqrt[3] x = \infty$ E. The function is decreasing on the interval $(-\infty, 1)$ The function is increasing on the interval $(1, \infty)$ F. The local minimum is $(1, -3)$ G. The graph is concave down on the interval $(-2,0)$ The graph is concave up on the intervals $(-\infty, -2)\cup (0,\infty)$ The points of inflection are $(-2, 7.56)$ and $(0,0)$ H. We can see a sketch of the curve below.

$y = (x-4)~\sqrt[3] x$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x = 0$, then $y = (0-4)~\sqrt[3] 0= 0$ The y-intercept is $0$ When $y = 0$: $(x-4)~\sqrt[3] x = 0$ $x = 0$ or $x=4$ The x-intercepts are $0$ and $4$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} (x-4)~\sqrt[3] x = \infty$ $\lim\limits_{x \to \infty} (x-4)~\sqrt[3] x = \infty$ There is no horizontal asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \sqrt[3] x+(x-4)(\frac{1}{3x^{2/3}})$ $y' = \frac{(3x)+(x-4)}{3x^{2/3}}$ $y' = \frac{4x-4}{3x^{2/3}} = 0$ $4x-4 = 0$ $x=1$ Note that $y'$ is undefined when $x=0$ When $x \lt 0$ or $0 \lt x \lt 1$ then $y' \lt 0$ The function is decreasing on the interval $(-\infty, 1)$ When $x \gt 1$, then $y' \gt 0$ The function is increasing on the interval $(1, \infty)$ F. When $x = 1$, then $y = (1-4)~\sqrt[3] 1= -3$ The local minimum is $(1, -3)$ G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{4(3x^{2/3})-(4x-4)(\frac{2}{x^{1/3}})}{9x^{4/3}}$ $y'' = \frac{4(3x)-(4x-4)(2)}{9x^{5/3}}$ $y'' = \frac{4x+8}{9x^{5/3}} = 0$ $4x+8 = 0$ $x = -2$ Note that $y''$ is undefined when $x=0$ When $-2 \lt x \lt 0$, then $y'' \lt 0$ The graph is concave down on the interval $(-2,0)$ When $x \lt -2$ or $x \gt 0$, then $y'' \gt 0$ The graph is concave up on the intervals $(-\infty, -2)\cup (0,\infty)$ When $x = -2$, then $y = (-2-4)~\sqrt[3] {-2}= 7.56$ When $x = 0$, then $y = (0-4)~\sqrt[3] 0= 0$ The points of inflection are $(-2, 7.56)$ and $(0,0)$ H. We can see a sketch of the curve below.