Calculus: Early Transcendentals 8th Edition

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