Solve x/x^3 <2 - ScanMath

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, - \frac{2}{2}) \cup (\frac{2}{2}, + \infty)

Evaluate

\frac{x}{x ^{3}} < 2

Find the domain

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\frac{x}{x ^{3}} < 2, x \neq = 0

Divide the terms

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

Move the expression to the left side

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

Subtract the terms

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

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

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

Calculate

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x = \frac{2}{2} x = - \frac{2}{2} x^{2} = 0

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

x = \frac{2}{2} x = - \frac{2}{2} x = 0

Determine the test intervals using the critical values

x < - \frac{2}{2} - \frac{2}{2} < x < 0 0 < x < \frac{2}{2} x > \frac{2}{2}

Choose a value form each interval

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

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

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x < - \frac{2}{2} is the solution x_{2} = - \frac{2}{4} x_{3} = \frac{2}{4} x_{4} = 2

To determine if - \frac{2}{2} < x < 0 is the solution to the inequality,test if the chosen value x = - \frac{2}{4} satisfies the initial inequality

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

To determine if 0 < x < \frac{2}{2} is the solution to the inequality,test if the chosen value x = \frac{2}{4} satisfies the initial inequality

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

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

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x < - \frac{2}{2} is the solution - \frac{2}{2} < x < 0 is not a solution 0 < x < \frac{2}{2} is not a solution x > \frac{2}{2} is the solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is x \in (- \infty, - \frac{2}{2}) \cup (\frac{2}{2}, + \infty)

x \in (- \infty, - \frac{2}{2}) \cup (\frac{2}{2}, + \infty)

Check if the solution is in the defined range

x \in (- \infty, - \frac{2}{2}) \cup (\frac{2}{2}, + \infty), x \neq = 0

Solution

x \in (- \infty, - \frac{2}{2}) \cup (\frac{2}{2}, + \infty)

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