## Calculus: Early Transcendentals 8th Edition

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