Answer
A. The domain is $[-1,0)\cup (0, 1]$
B. There is no y-intercept.
The x-intercepts are $-1, 1$
C. The function is an odd function.
D. $\lim\limits_{x \to 0^-} \frac{\sqrt{1-x^2}}{x} = -\infty$
$\lim\limits_{x \to 0^+} \frac{\sqrt{1-x^2}}{x} = \infty$
$x = 0$ is a vertical asymptote.
E. The graph is decreasing on the intervals $(-1,0)\cup (0,1)$
F. There is no local minimum or local maximum.
G. The graph is concave up on the intervals $(-1,-0.8165)\cup (0, 0.8165)$
The graph is concave down on the intervals $(-0.8165, 0)\cup (0.8165,1)$
The points of inflection are $(-0.8165,-0.707)\cup (0.8165, 0.707)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{\sqrt{1-x^2}}{x}$
A. The function is defined for real numbers such that $x \neq 0$ and $(1-x^2) \geq 0$:
$x^2 \leq 1$
$-1 \leq x \leq 1$
The domain is $[-1,0)\cup (0, 1]$
B. Since $x \neq 0$, there is no y-intercept.
When $y = 0$:
$\frac{\sqrt{1-x^2}}{x} = 0$
$\sqrt{1-x^2} = 0$
$x = -1, 1$
The x-intercepts are $-1, 1$
C. $\frac{\sqrt{1-(-x)^2}}{-x} = -\frac{\sqrt{1-x^2}}{x}$
The function is an odd function.
D. $\lim\limits_{x \to 0^-} \frac{\sqrt{1-x^2}}{x} = -\infty$
$\lim\limits_{x \to 0^+} \frac{\sqrt{1-x^2}}{x} = \infty$
$x = 0$ is a vertical asymptote.
E. We can find values of $x$ such that $y' = 0$:
$y' = \frac{(\frac{-x}{\sqrt{1-x^2}})(x)-(\sqrt{1-x^2})}{x^2}$
$y' = \frac{(-x^2)-(1-x^2)}{x^2~\sqrt{1-x^2}}$
$y' = \frac{-1}{x^2~\sqrt{1-x^2}}$
The are no values of $x$ such that $y' = 0$
When $-1 \lt x \lt 0$ or $0 \lt x \lt 1$, then $y' \lt 0$
The graph is decreasing on the intervals $(-1,0)\cup (0,1)$
F. There is no local minimum or local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{-(-1)(2x\sqrt{1-x^2}+\frac{-x^3}{\sqrt{1-x^2}})}{x^4~(1-x^2)}$
$y'' = \frac{2x(1-x^2)-x^3}{x^4~(1-x^2)^{3/2}}$
$y'' = \frac{2x-3x^3}{x^4~(1-x^2)^{3/2}}$
$y'' = \frac{2-3x^2}{x^3~(1-x^2)^{3/2}} = 0$
$2-3x^2 = 0$
$x^2 = \frac{2}{3}$
$x = \pm \sqrt{\frac{2}{3}}$
$x = -0.8165, 0.8165$
When $-1 \lt x \lt -0.8165$ or $0 \lt x \lt 0.8165$, then $y'' \gt 0$
The graph is concave up on the intervals $(-1,-0.8165)\cup (0, 0.8165)$
When $-0.8165 \lt x \lt 0$ or $0.8165 \lt x \lt 1$, then $y'' \lt 0$
The graph is concave down on the intervals $(-0.8165, 0)\cup (0.8165,1)$
When $x = -0.8165$, then $y = \frac{\sqrt{1-(-0.8165)^2}}{-0.8165} = -0.707$
When $x = 0.8165$, then $y = \frac{\sqrt{1-(0.8165)^2}}{0.8165} = 0.707$
The points of inflection are $(-0.8165,-0.707)\cup (0.8165, 0.707)$
H. We can see a sketch of the curve below.
