Solve 15(1/3b^3)<=6(b 9)

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

b \in (- \infty, - \frac{3 30}{5}] \cup [0, \frac{3 30}{5}]

Evaluate

15 \times \frac{1}{3} b^{3} \leq 6 (b \times 9)

Remove the parentheses

15 \times \frac{1}{3} b^{3} \leq 6 b \times 9

Multiply the numbers

More Steps

5 b^{3} \leq 6 b \times 9

Multiply the terms

5 b^{3} \leq 54 b

Move the expression to the left side

5 b^{3} - 54 b \leq 0

Rewrite the expression

5 b^{3} - 54 b = 0

Factor the expression

b (5 b^{2} - 54) = 0

Separate the equation into 2 possible cases

b = 0 5 b^{2} - 54 = 0

Solve the equation

More Steps

b = 0 b = \frac{3 30}{5} b = - \frac{3 30}{5}

Determine the test intervals using the critical values

b < - \frac{3 30}{5} - \frac{3 30}{5} < b < 0 0 < b < \frac{3 30}{5} b > \frac{3 30}{5}

Choose a value form each interval

b_{1} = - 4 b_{2} = - 2 b_{3} = 2 b_{4} = 4

To determine if b < - \frac{3 30}{5} is the solution to the inequality,test if the chosen value b = - 4 satisfies the initial inequality

More Steps

b < - \frac{3 30}{5} is the solution b_{2} = - 2 b_{3} = 2 b_{4} = 4

To determine if - \frac{3 30}{5} < b < 0 is the solution to the inequality,test if the chosen value b = - 2 satisfies the initial inequality

More Steps

b < - \frac{3 30}{5} is the solution - \frac{3 30}{5} < b < 0 is not a solution b_{3} = 2 b_{4} = 4

To determine if 0 < b < \frac{3 30}{5} is the solution to the inequality,test if the chosen value b = 2 satisfies the initial inequality

More Steps

b < - \frac{3 30}{5} is the solution - \frac{3 30}{5} < b < 0 is not a solution 0 < b < \frac{3 30}{5} is the solution b_{4} = 4

To determine if b > \frac{3 30}{5} is the solution to the inequality,test if the chosen value b = 4 satisfies the initial inequality

More Steps

b < - \frac{3 30}{5} is the solution - \frac{3 30}{5} < b < 0 is not a solution 0 < b < \frac{3 30}{5} is the solution b > \frac{3 30}{5} is not a solution

The original inequality is a nonstrict inequality,so include the critical value in the solution

b \leq - \frac{3 30}{5} is the solution 0 \leq b \leq \frac{3 30}{5} is the solution

Solution

b \in (- \infty, - \frac{3 30}{5}] \cup [0, \frac{3 30}{5}]

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