Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 321: 13

A. The domain is $(-\infty,-2) \cup (-2,2)\cup (2, \infty)$ B. The y-intercept is $0$ The x-intercept is $0$ C. The function is an odd function. D. $y = 0$ is a horizontal asymptote. $x = -2$ is a vertical asymptote. $x = 2$ is a vertical asymptote. E. The function is decreasing on the intervals $(-\infty, -2)\cup (-2,2)\cup (2,\infty)$ F. There is no local local minimum or local maximum. G. The graph is concave up on the intervals $(-2,0)\cup (2,\infty)$ The graph is concave down on the intervals $(-\infty, -2)\cup (0,2)$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.

$y = \frac{x}{x^2-4}$ A. The function is defined for all real numbers except $x = -2,2$ The domain is $(-\infty,-2) \cup (-2,2)\cup (2, \infty)$ B. When $x = 0$, then $y = \frac{0}{0^2-4} = 0$ The y-intercept is $0$ When $y = 0$: $\frac{x}{x^2-4} = 0$ $x = 0$ The x-intercept is $0$ C. $\frac{(-x)}{(-x)^2-4} = -\frac{x}{x^2-4}$ The function is an odd function. D. $\lim\limits_{x \to -\infty} (\frac{x}{x^2-4}) = 0$ $\lim\limits_{x \to \infty} (\frac{x}{x^2-4}) = 0$ $y = 0$ is a horizontal asymptote. $\lim\limits_{x \to -2^-} (\frac{x}{x^2-4}) = -\infty$ $\lim\limits_{x \to -2^+} (\frac{x}{x^2-4}) = \infty$ $x = -2$ is a vertical asymptote. $\lim\limits_{x \to 2^-} (\frac{x}{x^2-4}) = -\infty$ $\lim\limits_{x \to 2^+} (\frac{x}{x^2-4}) = \infty$ $x = 2$ is a vertical asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \frac{(x^2-4)-(x)(2x)}{(x^2-4)^2}$ $y' = \frac{-x^2-4}{(x^2-4)^2} = 0$ $-x^2-4 = 0$ $x^2+4 = 0$ There are no values of $x$ such that $y' = 0$ When $x \lt -2$ or $-2 \lt x \lt 2$ or $x \gt 2$, then $y' \lt 0$ The function is decreasing on the intervals $(-\infty, -2)\cup (-2,2)\cup (2,\infty)$ F. There is no local local minimum or local maximum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{(-2x)(x^2-4)^2-(-x^2-4)(2)(x^2-4)(2x)}{(x^2-4)^4}$ $y'' = \frac{(-2x)(x^2-4)-(-x^2-4)(2)(2x)}{(x^2-4)^3}$ $y'' = \frac{-2x^3+8x+4x^3+16x}{(x^2-4)^3}$ $y'' = \frac{2x^3+24x}{(x^2-4)^3} = 0$ $2x^3+24x = 0$ $2x(x^2+12) = 0$ $x = 0$ When $-2 \lt x \lt 0~~$ or $x \gt 2$, then $y'' \gt 0$ The graph is concave up on the intervals $(-2,0)\cup (2,\infty)$ When $x \lt -2$ or $0 \lt x \lt 2$, then $y'' \lt 0$ The graph is concave down on the intervals $(-\infty, -2)\cup (0,2)$ When $x = 0$, then $y = \frac{0}{0^2-4} = 0$ The point of inflection is $(0,0)$ H. We can see a sketch of the curve below.