Question

= \frac{x^2 - 1}{x^3}
ype dne if the answer does not exist for any of the characteristics below

• domain

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• intercepts
• x - intercept(s)

if the are multiple points, separate with a comma. ex. (-9,0),(4,0)
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• y - intercept

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• symmetry

select an answer dropdown

• asymptotes
• horizontal asymptote

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• vertical asymptote

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• intervals of increase and decrease
• interval of increase

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• interval of decease

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• local max and/or min
• list all maximum points

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• list all minimum points

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• concavity and inflection points
• interval of concave up

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Explanation:

Step1: Find the Domain

The function is \( f(x)=\frac{x^{2}-1}{x^{3}} \). The denominator cannot be zero, so \( x^{3} eq0\Rightarrow x eq0 \). Thus, the domain is all real numbers except \( x = 0 \), which can be written as \( (-\infty,0)\cup(0,\infty) \).

Step2: Find \( x \)-intercepts

Set \( y = 0 \), so \( \frac{x^{2}-1}{x^{3}}=0 \). The numerator must be zero (denominator non - zero), \( x^{2}-1 = 0\Rightarrow(x - 1)(x + 1)=0\Rightarrow x=1\) or \( x=-1 \). So the \( x \)-intercepts are \( (-1,0),(1,0) \).

Step3: Find \( y \)-intercept

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

Step4: Check Symmetry

A function \( f(x) \) is odd if \( f(-x)=-f(x) \), even if \( f(-x)=f(x) \).
\( f(-x)=\frac{(-x)^{2}-1}{(-x)^{3}}=\frac{x^{2}-1}{-x^{3}}=-\frac{x^{2}-1}{x^{3}}=-f(x) \). So the function is odd, symmetric about the origin.

Step5: Find Asymptotes

- **Horizontal Asymptote**:
We look at the degrees of the numerator and denominator. The degree of the numerator \( n = 2 \), the degree of the denominator \( m = 3 \). Since \( n<m \), the horizontal asymptote is \( y = 0 \).
- **Vertical Asymptote**:
The denominator is zero at \( x = 0 \) and the numerator is non - zero at \( x = 0 \) (\( 0^{2}-1=-1 eq0 \)), so the vertical asymptote is \( x = 0 \).

Step6: Find Intervals of Increase/Decrease (First, find the derivative)

First, find the derivative of \( f(x)=\frac{x^{2}-1}{x^{3}}=x^{-1}-x^{-3} \) using the power rule.
\( f^\prime(x)=-x^{-2}+3x^{-4}=\frac{-1}{x^{2}}+\frac{3}{x^{4}}=\frac{-x^{2}+3}{x^{4}}=\frac{-(x^{2}-3)}{x^{4}}=\frac{- (x-\sqrt{3})(x + \sqrt{3})}{x^{4}} \)
The critical points are where \( f^\prime(x)=0 \) or undefined. \( f^\prime(x) \) is undefined at \( x = 0 \). \( f^\prime(x)=0\Rightarrow x^{2}-3 = 0\Rightarrow x=\pm\sqrt{3} \)
We test intervals:
- For \( x\in(-\infty,-\sqrt{3}) \), let's take \( x=-2 \), \( f^\prime(-2)=\frac{-((-2)^{2}-3)}{(-2)^{4}}=\frac{-(4 - 3)}{16}=\frac{-1}{16}<0 \), so the function is decreasing.
- For \( x\in(-\sqrt{3},0) \), let's take \( x = - 1 \), \( f^\prime(-1)=\frac{-((-1)^{2}-3)}{(-1)^{4}}=\frac{- (1 - 3)}{1}=\ 2>0 \), so the function is increasing.
- For \( x\in(0,\sqrt{3}) \), let's take \( x = 1 \), \( f^\prime(1)=\frac{- (1 - 3)}{1}=2>0 \), so the function is increasing.
- For \( x\in(\sqrt{3},\infty) \), let's take \( x = 2 \), \( f^\prime(2)=\frac{- (4 - 3)}{16}=\frac{-1}{16}<0 \), so the function is decreasing.

So the interval of increase is \( (-\sqrt{3},0)\cup(0,\sqrt{3}) \) and the interval of decrease is \( (-\infty,-\sqrt{3})\cup(\sqrt{3},\infty) \)

Step7: Find Local Max/Min

Using the first derivative test:
- At \( x = -\sqrt{3} \), the function changes from decreasing to increasing, so there is a local minimum. \( f(-\sqrt{3})=\frac{(-\sqrt{3})^{2}-1}{(-\sqrt{3})^{3}}=\frac{3 - 1}{-3\sqrt{3}}=\frac{2}{-3\sqrt{3}}=-\frac{2\sqrt{3}}{9} \). So the local minimum point is \( (-\sqrt{3},-\frac{2\sqrt{3}}{9}) \)
- At \( x=\sqrt{3} \), the function changes from increasing to decreasing, so there is a local maximum. \( f(\sqrt{3})=\frac{(\sqrt{3})^{2}-1}{(\sqrt{3})^{3}}=\frac{3 - 1}{3\sqrt{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9} \). So the local maximum point is \( (\sqrt{3},\frac{2\sqrt{3}}{9}) \)

Step8: Find Concavity and Inflection Points (Find the second derivative)

First, \( f^\prime(x)=\frac{-x^{2}+3}{x^{4}} \)
Using the quotient rule \( (u/v)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}} \), where \( u=-x^{2}+3 \), \( u^\prime=-2x \), \( v = x^{4} \), \( v^\prime = 4x^{3} \)
\( f^{\prime\prime}(x)=\frac{(-2x)x^{4}-(-x^{2}+3)(4x^{3})}{(x^{4})^{2}}=\frac{-2x^{5}+4x^{5}-12x^{3}}{x^{8}}=\frac{2x^{5}-12x^{3}}{x^{8}}=\frac{2x^{3}(x^{2}-6)}{x^{8}}=\frac{2(x^{2}-6)}{x^{5}}=\frac{2(x-\sqrt{6})(x + \sqrt{6})}{x^{5}} \)
The critical points for concavity are where \( f^{\prime\prime}(x)=0 \) or undefined. \( f^{\prime\prime}(x) \) is undefined at \( x = 0 \). \( f^{\prime\prime}(x)=0\Rightarrow x^{2}-6 = 0\Rightarrow x=\pm\sqrt{6} \)
Test intervals:
- For \( x\in(-\infty,-\sqrt{6}) \), let \( x=-3 \), \( f^{\prime\prime}(-3)=\frac{2((-3)^{2}-6)}{(-3)^{5}}=\frac{2(9 - 6)}{-243}=\frac{6}{-243}<0 \), concave down.
- For \( x\in(-\sqrt{6},0) \), let \( x=-2 \), \( f^{\prime\prime}(-2)=\frac{2(4 - 6)}{(-2)^{5}}=\frac{2(-2)}{-32}=\frac{-4}{-32}>0 \), concave up.
- For \( x\in(0,\sqrt{6}) \), let \( x = 2 \), \( f^{\prime\prime}(2)=\frac{2(4 - 6)}{2^{5}}=\frac{2(-2)}{32}=\frac{-4}{32}<0 \), concave down.
- For \( x\in(\sqrt{6},\infty) \), let \( x = 3 \), \( f^{\prime\prime}(3)=\frac{2(9 - 6)}{3^{5}}=\frac{6}{243}>0 \), concave up.

Inflection points occur where concavity changes. At \( x = -\sqrt{6} \), \( f(-\sqrt{6})=\frac{(-\sqrt{6})^{2}-1}{(-\sqrt{6})^{3}}=\frac{6 - 1}{-6\sqrt{6}}=\frac{5}{-6\sqrt{6}}=-\frac{5\sqrt{6}}{36} \)
At \( x=\sqrt{6} \), \( f(\sqrt{6})=\frac{(\sqrt{6})^{2}-1}{(\sqrt{6})^{3}}=\frac{6 - 1}{6\sqrt{6}}=\frac{5}{6\sqrt{6}}=\frac{5\sqrt{6}}{36} \)
So the intervals of concave up are \( (-\sqrt{6},0)\cup(\sqrt{6},\infty) \) and concave down are \( (-\infty,-\sqrt{6})\cup(0,\sqrt{6}) \), inflection points are \( (-\sqrt{6},-\frac{5\sqrt{6}}{36}),(\sqrt{6},\frac{5\sqrt{6}}{36}) \)

(We can present the answers for each part as per the requirements of the problem. Here we have covered all the main parts of analyzing the function \( y=\frac{x^{2}-1}{x^{3}} \))

Answer:

• Domain: \( (-\infty,0)\cup(0,\infty) \)
• \( x \)-intercept(s): \( (-1,0),(1,0) \)
• \( y \)-intercept: DNE
• Symmetry: Symmetric about the origin (Odd function)
• Horizontal Asymptote: \( y = 0 \)
• Vertical Asymptote: \( x = 0 \)
• Interval of Increase: \( (-\sqrt{3},0)\cup(0,\sqrt{3}) \)
• Interval of Decrease: \( (-\infty,-\sqrt{3})\cup(\sqrt{3},\infty) \)
• Local maximum point: \( (\sqrt{3},\frac{2\sqrt{3}}{9}) \)
• Local minimum point: \( (-\sqrt{3},-\frac{2\sqrt{3}}{9}) \)
• Interval of Concave up: \( (-\sqrt{6},0)\cup(\sqrt{6},\infty) \)
• Interval of Concave down: \( (-\infty,-\sqrt{6})\cup(0,\sqrt{6}) \)
• Inflection Points: \( (-\sqrt{6},-\frac{5\sqrt{6}}{36}),(\sqrt{6},\frac{5\sqrt{6}}{36}) \)