9x-x^3>= 0

Solution

9 x - x^{3} \geq 0

x \leq - 3 or 0 \leq x \leq 3

Interval Notation

(- \infty, - 3] \cup [0, 3]

Solution steps

9 x - x^{3} \geq 0

Factor

9 x - x^{3} : - x (x + 3) (x - 3)

- x (x + 3) (x - 3) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x + 3) (x - 3)) (- 1) \leq 0 \cdot (- 1)

Simplify

x (x + 3) (x - 3) \leq 0

Identify the intervals

x \leq - 3 or 0 \leq x \leq 3

k^{2} = 19

(x + 6)^{2} = 49

y^{2} - 4 = 0

4 x + 6 < 2 x + 1

2 x^{2} < 7 x + 15