Solve (7x-5)/(8x^3) >4

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

- 0.671811 < x < 0

Alternative Form

x \in (- 0.671811, 0)

Evaluate

\frac{7 x - 5}{8 x ^{3}} > 4

Find the domain

More Steps

\frac{7 x - 5}{8 x ^{3}} > 4, x \neq = 0

Move the expression to the left side

\frac{7 x - 5}{8 x ^{3}} - 4 > 0

Subtract the terms

More Steps

\frac{7 x - 5 - 32 x ^{3}}{8 x ^{3}} > 0

Set the numerator and denominator of \frac{7 x - 5 - 32 x ^{3}}{8 x ^{3}} equal to 0 to find the values of x where sign changes may occur

7 x - 5 - 32 x^{3} = 0 8 x^{3} = 0

Calculate

x \approx - 0.671811 8 x^{3} = 0

Calculate

More Steps

x \approx - 0.671811 x = 0

Determine the test intervals using the critical values

x < - 0.671811 - 0.671811 < x < 0 x > 0

Choose a value form each interval

x_{1} = - 2 x_{2} \approx - 0.335906 x_{3} = 1

To determine if x < - 0.671811 is the solution to the inequality,test if the chosen value x = - 2 satisfies the initial inequality

More Steps

x < - 0.671811 is not a solution x_{2} \approx - 0.335906 x_{3} = 1

To determine if - 0.671811 < x < 0 is the solution to the inequality,test if the chosen value x \approx - 0.335906 satisfies the initial inequality

More Steps

x < - 0.671811 is not a solution - 0.671811 < x < 0 is the solution x_{3} = 1

To determine if x > 0 is the solution to the inequality,test if the chosen value x = 1 satisfies the initial inequality

More Steps

x < - 0.671811 is not a solution - 0.671811 < x < 0 is the solution x > 0 is not a solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is - 0.671811 < x < 0

- 0.671811 < x < 0

Check if the solution is in the defined range

- 0.671811 < x < 0, x \neq = 0

Solution

- 0.671811 < x < 0

Alternative Form

x \in (- 0.671811, 0)

Show Solution