Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 322: 20

A. The domain is $(-\infty,2)\cup (2, \infty)$ B. The y-intercept is $0$ The x-intercept is $0$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \frac{x^3}{x-2} = \infty$ $\lim\limits_{x \to \infty} \frac{x^3}{x-2} = \infty$ There is no horizontal asymptote. $\lim\limits_{x \to 2^-} \frac{x^3}{x-2} = -\infty$ $\lim\limits_{x \to 2^+} \frac{x^3}{x-2} = \infty$ $x = 2$ is a vertical asymptote. E. The function is decreasing on the intervals $(-\infty, 0)\cup (0,2)\cup (2,3)$ The function is increasing on the intervals $(3, \infty)$ F. The local minimum is $(3, 27)$ G. The graph is concave down on the interval $(0,2)$ The graph is concave up on the intervals $(-\infty,0)\cup (2,\infty)$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.

$y = \frac{x^3}{x-2}$ A. The function is defined for all real numbers except $x=2$ The domain is $(-\infty,2)\cup (2, \infty)$ B. When $x = 0$, then $y = \frac{0^3}{0-2}= 0$ The y-intercept is $0$ When $y = 0$: $\frac{x^3}{x-2} = 0$ $x^3 = 0$ $x = 0$ The x-intercept is $0$ C. The function is not an even function or an odd function. D. $\lim\limits_{x \to -\infty} \frac{x^3}{x-2} = \infty$ $\lim\limits_{x \to \infty} \frac{x^3}{x-2} = \infty$ There is no horizontal asymptote. $\lim\limits_{x \to 2^-} \frac{x^3}{x-2} = -\infty$ $\lim\limits_{x \to 2^+} \frac{x^3}{x-2} = \infty$ $x = 2$ is a vertical asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \frac{(3x^2)(x-2)-(x^3)(1)}{(x-2)^2} = \frac{2x^3-6x^2}{(x-2)^2} = 0$ $2x^3-6x^2 = 0$ $2x^2(x-3) = 0$ $x = 0, 3$ When $x \lt 0$ or $0 \lt x \lt 2$ or $2 \lt x \lt 3$, then $y' \lt 0$ The function is decreasing on the intervals $(-\infty, 0)\cup (0,2)\cup (2,3)$ When $x \gt 3$, then $y' \gt 0$ The function is increasing on the intervals $(3, \infty)$ F. When $x = 3$, then $y = \frac{3^3}{3-2}= 27$ The local minimum is $(3, 27)$ G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{(6x^2-12x)(x-2)^2-(2x^3-6x^2)(2)(x-2)}{(x-2)^4}$ $y'' = \frac{(6x^2-12x)(x-2)-(2x^3-6x^2)(2)}{(x-2)^3}$ $y'' = \frac{6x^3-24x^2+24x-4x^3+12x^2}{(x-2)^3}$ $y'' = \frac{2x^3-12x^2+24x}{(x-2)^3} = 0$ $2x^3-12x^2+24x = 0$ $2x(x^2-6x+12) = 0$ $x = 0$ When $0 \lt x \lt 2$ then $y'' \lt 0$ The graph is concave down on the interval $(0,2)$ When $x \lt 0$ or $x \gt 2$, then $y'' \gt 0$ The graph is concave up on the intervals $(-\infty,0)\cup (2,\infty)$ When $x = 0$, then $y = \frac{0^3}{0-2}= 0$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.