Answer
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.
Work Step by Step
$y = \frac{x}{\sqrt{x^2+1}}$
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.
