Question

y = \frac{x^{2}-1}{x^{3}}
enter dne if an answer does not exist.

• domain

• intercepts

o x - intercept(s)
separate multiple points with a comma, e.g., (-9,0),(4,0).
o y - intercept

• symmetry

select an answer

• asymptotes

o horizontal asymptote
o vertical asymptote

• intervals of increase and decrease

o interval of increase
o interval of decease

• local max and/or min

o list all maximum points
o list all minimum points

• concavity and inflection points

o interval of concave up
o interval of concave down
o inflections point(s)
separate points with a comma.

• graph

using the information above, sketch the given function and then identify the choices displayed below.

Explanation:

Step1: Find the domain

The function is $y = \frac{x^{2}-1}{x^{3}}$. The denominator cannot be zero. So, $x
eq0$. The domain is $(-\infty,0)\cup(0,\infty)$.

Step2: Find the x - intercepts

Set $y = 0$. Then $\frac{x^{2}-1}{x^{3}}=0$, which means $x^{2}-1 = 0$ and $x
eq0$. Solving $x^{2}-1=0$ gives $x=- 1,1$. So the x - intercepts are $(-1,0),(1,0)$.

Step3: Find the y - intercept

Set $x = 0$, but the function is undefined at $x = 0$, so there is no y - intercept (DNE).

Step4: Check for symmetry

Replace $x$ with $-x$: $y=\frac{(-x)^{2}-1}{(-x)^{3}}=-\frac{x^{2}-1}{x^{3}}$, so the function is odd and symmetric about the origin.

Step5: Find the horizontal asymptote

$\lim_{x
ightarrow\pm\infty}\frac{x^{2}-1}{x^{3}}=\lim_{x
ightarrow\pm\infty}\frac{\frac{x^{2}}{x^{3}}-\frac{1}{x^{3}}}{1}=\lim_{x
ightarrow\pm\infty}(\frac{1}{x}-\frac{1}{x^{3}})=0$. So the horizontal asymptote is $y = 0$.

Step6: Find the vertical asymptote

Since the function is undefined at $x = 0$, the vertical asymptote is $x = 0$.

Step7: Find the derivative

Using the quotient rule, if $y=\frac{u}{v}$ where $u=x^{2}-1$ and $v = x^{3}$, then $y'=\frac{u'v - uv'}{v^{2}}$. $u'=2x$ and $v'=3x^{2}$. So $y'=\frac{2x\cdot x^{3}-(x^{2}-1)\cdot3x^{2}}{x^{6}}=\frac{2x^{4}-3x^{4}+3x^{2}}{x^{6}}=\frac{-x^{4}+3x^{2}}{x^{6}}=\frac{-x^{2}(x^{2} - 3)}{x^{6}}=\frac{-(x^{2}-3)}{x^{4}}=\frac{-x^{2}+3}{x^{4}}$. Set $y'=0$, then $-x^{2}+3 = 0$, so $x=\pm\sqrt{3}$.

• The function is increasing on $(-\sqrt{3},0)$ and $(0,\sqrt{3})$ since $y'>0$ on these intervals.
• The function is decreasing on $(-\infty,-\sqrt{3})$ and $(\sqrt{3},\infty)$ since $y'<0$ on these intervals.

Step8: Find local maxima and minima

Since the function changes from increasing to decreasing at $x=\sqrt{3}$, the local maximum is $(\sqrt{3},\frac{3 - 1}{3\sqrt{3}})=(\sqrt{3},\frac{2}{3\sqrt{3}})$. Since the function changes from increasing to decreasing at $x =-\sqrt{3}$, the local maximum is $(-\sqrt{3},\frac{3 - 1}{- 3\sqrt{3}})=(-\sqrt{3},-\frac{2}{3\sqrt{3}})$.

Step9: Find the second - derivative

$y'=\frac{-x^{2}+3}{x^{4}}$, using the quotient rule again. Let $u=-x^{2}+3$ and $v = x^{4}$, $u'=-2x$ and $v'=4x^{3}$. Then $y''=\frac{-2x\cdot x^{4}-(-x^{2}+3)\cdot4x^{3}}{x^{8}}=\frac{-2x^{5}+4x^{5}-12x^{3}}{x^{8}}=\frac{2x^{5}-12x^{3}}{x^{8}}=\frac{2x^{3}(x^{2}-6)}{x^{8}}=\frac{2(x^{2}-6)}{x^{5}}$.

• Set $y'' = 0$, then $x^{2}-6=0$, so $x=\pm\sqrt{6}$.
• The function is concave up on $(-\sqrt{6},0)$ and $(\sqrt{6},\infty)$ since $y''>0$ on these intervals.
• The function is concave down on $(-\infty,-\sqrt{6})$ and $(0,\sqrt{6})$ since $y''<0$ on these intervals.
• The inflection points are $(-\sqrt{6},-\frac{6 - 1}{6\sqrt{6}})=(-\sqrt{6},-\frac{5}{6\sqrt{6}})$ and $(\sqrt{6},\frac{6 - 1}{6\sqrt{6}})=(\sqrt{6},\frac{5}{6\sqrt{6}})$.

Answer:

• Domain: $(-\infty,0)\cup(0,\infty)$
• x - intercept(s): $(-1,0),(1,0)$
• y - intercept: DNE
• Symmetry: Origin
• Horizontal Asymptote: $y = 0$
• Vertical Asymptote: $x = 0$
• Interval of Increase: $(-\sqrt{3},0),(0,\sqrt{3})$
• Interval of Decrease: $(-\infty,-\sqrt{3}),(\sqrt{3},\infty)$
• List all maximum points: $(\sqrt{3},\frac{2}{3\sqrt{3}}),(-\sqrt{3},-\frac{2}{3\sqrt{3}})$
• List all minimum points: None
• Interval of Concave up: $(-\sqrt{6},0),(\sqrt{6},\infty)$
• Interval of Concave Down: $(-\infty,-\sqrt{6}),(0,\sqrt{6})$
• Inflections Point(s): $(-\sqrt{6},-\frac{5}{6\sqrt{6}}),(\sqrt{6},\frac{5}{6\sqrt{6}})$