## Calculus: Early Transcendentals 8th Edition

A. The domain is $(-\infty,-2) \cup (-2,2)\cup (2, \infty)$ B. The y-intercept is $-\frac{1}{4}$ There are no x-intercepts. C. The function is an even 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 $(0,2)\cup (2,\infty)$ The function is increasing on the intervals $(-\infty, -2)\cup (-2,0)$ F. The local maximum is $(0, -\frac{1}{4})$ G. The graph is concave up on the intervals $(-\infty,-2)\cup (2,\infty)$ The graph is concave down on the intervals $(-2,2)$ There are no points of inflection. H. We can see a sketch of the curve below.
$y = \frac{1}{x^2-4}$ 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 $y = \frac{1}{0^2-4} = -\frac{1}{4}$ The y-intercept is $-\frac{1}{4}$ When $y = 0$: $\frac{1}{x^2-4} = 0$ There are no values of $x$ such that $y = 0$ There are no x-intercepts. C. $\frac{1}{(-x)^2-4} = \frac{1}{x^2-4}$ The function is an even function. D. $\lim\limits_{x \to -\infty} (\frac{1}{x^2-4}) = 0$ $\lim\limits_{x \to \infty} (\frac{1}{x^2-4}) = 0$ $y = 0$ is a horizontal asymptote. $\lim\limits_{x \to -2^-} (\frac{1}{x^2-4}) = \infty$ $\lim\limits_{x \to -2^+} (\frac{1}{x^2-4}) = -\infty$ $x = -2$ is a vertical asymptote. $\lim\limits_{x \to 2^-} (\frac{1}{x^2-4}) = -\infty$ $\lim\limits_{x \to 2^+} (\frac{1}{x^2-4}) = \infty$ $x = 2$ is a vertical asymptote. E. We can find the values of $x$ such that $y' = 0$: $y' = \frac{-(1)(2x)}{(x^2-4)^2} = \frac{-2x}{(x^2-4)^2} = 0$ $-2x = 0$ $x = 0$ When $0 \lt x \lt 2$ or $x \gt 2$, then $y' \lt 0$ The function is decreasing on the intervals $(0,2)\cup (2,\infty)$ When $x \lt -2$ or $-2 \lt x \lt 0$, then $y' \gt 0$ The function is increasing on the intervals $(-\infty, -2)\cup (-2,0)$ F. When $x = 0,$ then $y = \frac{1}{0^2-4} = -\frac{1}{4}$ The local maximum is $(0, -\frac{1}{4})$ G. We can find the values of $x$ such that $y'' = 0$: $y'' = \frac{(-2)(x^2-4)^2-(-2x)(2)(x^2-4)(2x)}{(x^2-4)^4}$ $y'' = \frac{-2x^2+8+8x^2}{(x^2-4)^3}$ $y'' = \frac{6x^2+8}{(x^2-4)^3} = 0$ $6x^2+8 = 0$ There are no values of $x$ such that $y'' = 0$ When $x \lt -2~~$ or $x \gt 2$, then $y'' \gt 0$ The graph is concave up on the intervals $(-\infty,-2)\cup (2,\infty)$ When $-2 \lt x \lt 2$, then $y'' \lt 0$ The graph is concave down on the intervals $(-2,2)$ There are no points of inflection. H. We can see a sketch of the curve below.