Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 30

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $0$ The x-intercepts are $0$ and $5$ C. The function is not an odd function or an even function. D. $\lim\limits_{x \to -\infty} (x^{5/3}-5x^{2/3}) = -\infty$ $\lim\limits_{x \to \infty} (x^{5/3}-5x^{2/3}) = \infty$ No asymptotes. E. The function is decreasing on the interval $(0,2)$ The function is increasing on the intervals $(-\infty, 0)\cup (2, \infty)$ F. The local minimum is $(2, -4.76)$ G. The graph is concave down on the interval $(-\infty,-1)$ The graph is concave up on the intervals $(-1,0) \cup (0, \infty)$ The point of inflection is $(-1,-6)$ H. We can see a sketch of the curve below.

$y = x^{5/3}-5x^{2/3}$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x=0$, then $y = 0^{5/3}-5(0)^{2/3} = 0$ The y-intercept is $0$ When $y = 0$: $x^{5/3}-5x^{2/3} = 0$ $x^{5/3}=5x^{2/3}$ $x = 0$ or $x = 5$ The x-intercepts are $0$ and $5$ C. The function is not an odd function or an even function. D. $\lim\limits_{x \to -\infty} (x^{5/3}-5x^{2/3}) = -\infty$ $\lim\limits_{x \to \infty} (x^{5/3}-5x^{2/3}) = \infty$ There are no horizontal asymptotes. E. We can find values of $x$ such that $y' = 0$: $y' = \frac{5}{3}\cdot x^{2/3}-\frac{10}{3}\cdot \frac{1}{x^{1/3}} = 0$ $\frac{5}{3}\cdot x^{2/3} = \frac{10}{3}\cdot \frac{1}{x^{1/3}}$ $x = 2$ Note that $y'$ is undefined when $x=0$ When $0 \lt x \lt 2$, then $y' \lt 0$ The function is decreasing on the interval $(0,2)$ When $x \lt 0$ or $x \gt 2$, then $y' \gt 0$ The function is increasing on the intervals $(-\infty, 0)\cup (2, \infty)$ F. When $x=2$, then $y = 2^{5/3}-5(2)^{2/3} = -4.76$ The local minimum is $(2, -4.76)$ G. We can find the values of $x$ such that $y'' = 0$: $y'' =\frac{10}{9}\cdot \frac{1}{x^{1/3}}+\frac{10}{9}\cdot \frac{1}{x^{4/3}} = 0$ $\frac{10}{9}\cdot \frac{1}{x^{1/3}}=-\frac{10}{9}\cdot \frac{1}{x^{4/3}}$ $x =-1$ Note that $y''$ is undefined when $x=0$ When $x \lt -1$, then $y'' \lt 0$ The graph is concave down on the interval $(-\infty,-1)$ When $-1 \lt x \lt 0$ or $x \gt 0$, then $y'' \gt 0$ The graph is concave up on the intervals $(-1,0) \cup (0, \infty)$ When $x=-1$, then $y = (-1)^{5/3}-5(-1)^{2/3} = -6$ The point of inflection is $(-1,-6)$ H. We can see a sketch of the curve below.