Solve x/(x^3) > (x^3)/x

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- 1, 0) \cup (0, 1)

Evaluate

\frac{x}{x ^{3}} > \frac{x ^{3}}{x}

Find the domain

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\frac{x}{x ^{3}} > \frac{x ^{3}}{x}, x \neq = 0

Divide the terms

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\frac{1}{x ^{2}} > \frac{x ^{3}}{x}

Divide the terms

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\frac{1}{x ^{2}} > x^{2}

Move the expression to the left side

\frac{1}{x ^{2}} - x^{2} > 0

Subtract the terms

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\frac{1 - x ^{4}}{x ^{2}} > 0

Set the numerator and denominator of \frac{1 - x ^{4}}{x ^{2}} equal to 0 to find the values of x where sign changes may occur

1 - x^{4} = 0 x^{2} = 0

Calculate

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x = 1 x = - 1 x^{2} = 0

The only way a power can be 0 is when the base equals 0

x = 1 x = - 1 x = 0

Determine the test intervals using the critical values

x < - 1 - 1 < x < 0 0 < x < 1 x > 1

Choose a value form each interval

x_{1} = - 2 x_{2} = - \frac{1}{2} x_{3} = \frac{1}{2} x_{4} = 2

To determine if x < - 1 is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

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x < - 1 is not a solution x_{2} = - \frac{1}{2} x_{3} = \frac{1}{2} x_{4} = 2

To determine if - 1 < x < 0 is the solution to the inequality,test if the chosen value x = - \frac{1}{2} satisfies the initial inequality

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x < - 1 is not a solution - 1 < x < 0 is the solution x_{3} = \frac{1}{2} x_{4} = 2

To determine if 0 < x < 1 is the solution to the inequality,test if the chosen value x = \frac{1}{2} satisfies the initial inequality

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x < - 1 is not a solution - 1 < x < 0 is the solution 0 < x < 1 is the solution x_{4} = 2

To determine if x > 1 is the solution to the inequality,test if the chosen value x = 2 satisfies the initial inequality

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x < - 1 is not a solution - 1 < x < 0 is the solution 0 < x < 1 is the solution x > 1 is not a solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is x \in (- 1, 0) \cup (0, 1)

x \in (- 1, 0) \cup (0, 1)

Check if the solution is in the defined range

x \in (- 1, 0) \cup (0, 1), x \neq = 0

Solution

x \in (- 1, 0) \cup (0, 1)

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