## Calculus: Early Transcendentals 8th Edition

A. The domain is $(-\infty,-1) \cup (1, \infty)$ B. There is no y-intercept. There are no x-intercepts. C. The function is an odd function. D. $y = -1$ is a horizontal asymptote. $y = 1$ is a horizontal asymptote. $x = -1$ is a vertical asymptote. $x = 1$ is a vertical asymptote. E. The function is decreasing on the intervals $(-\infty,-1)\cup (1, \infty)$ F. There is no local minimum or local maximum. G. The graph is concave down on the interval $(-\infty,-1)$ The graph is concave up on the interval $(1, \infty)$ There is no point of inflection. H. We can see a sketch of the curve below.
$y = \frac{x}{\sqrt{x^2-1}}$ A. The function is defined for all real numbers such that $x^2-1 \gt 0$ $x^2 \gt 1$ $x \lt -1$ or $x \gt 1$ The domain is $(-\infty,-1) \cup (1, \infty)$ B. Since $x \neq 0$, there is no y-intercept. When $y = 0$: $\frac{x}{\sqrt{x^2+1}} = 0$ There are no values of $x$ in the domain such that $y = 0$ There are no x-intercepts. C. $\frac{-x}{\sqrt{(-x)^2-1}} = -\frac{x}{\sqrt{x^2-1}}$ The function is an odd function. D. $\lim\limits_{x \to -\infty} \frac{x}{\sqrt{x^2-1}} = -1$ $y = -1$ is a horizontal asymptote. $\lim\limits_{x \to \infty} \frac{x}{\sqrt{x^2-1}} = 1$ $y = 1$ is a horizontal asymptote. $\lim\limits_{x \to -1^-} \frac{x}{\sqrt{x^2-1}} = -\infty$ $x = -1$ is a vertical asymptote. $\lim\limits_{x \to 1^+} \frac{x}{\sqrt{x^2-1}} = \infty$ $x = 1$ is a vertical asymptote. E. We can try to find values of $x$ such that $y' = 0$: $y' =\frac{\sqrt{x^2-1}-(x)(\frac{x}{\sqrt{x^2-1}})}{x^2-1}$ $y' =\frac{x^2-1-x^2}{(x^2-1)^{3/2}}$ $y' =\frac{-1}{(x^2-1)^{3/2}}$ There are no values of $x$ such that $y' = 0$ $y' \lt 0$ for all values of $x$ in the domain. The function is decreasing on the intervals $(-\infty,-1)\cup (1, \infty)$ F. There is no local minimum or local maximum. G. We can try to find the values of $x$ such that $y'' = 0$: $y'' = \frac{-(-1)(\frac{3x}{\sqrt{x^2-1}})}{(x^2-1)^3}$ $y'' = \frac{3x}{(x^2-1)^{5/2}}$ There are no values of $x$ in the domain such that $y' = 0$ When $x \lt -1$, then $y'' \lt 0$ The graph is concave down on the interval $(-\infty,-1)$ When $x \gt 1$, then $y'' \gt 0$ The graph is concave up on the interval $(1, \infty)$ There is no point of inflection. H. We can see a sketch of the curve below.