36x<= x^3

Solution

36 x \leq x^{3}

- 6 \leq x \leq 0 or x \geq 6

Interval Notation

[- 6, 0] \cup [6, \infty)

Solution steps

36 x \leq x^{3}

Rewrite in standard form

36 x - x^{3} \leq 0

Factor

36 x - x^{3} : - x (x + 6) (x - 6)

- x (x + 6) (x - 6) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (x + 6) (x - 6)) (- 1) \geq 0 \cdot (- 1)

Simplify

x (x + 6) (x - 6) \geq 0

Identify the intervals

- 6 \leq x \leq 0 or x \geq 6

\frac{1}{3 ^{- 5}}

2 + z = 6 z^{2}

- 17 = \frac{x}{5}

12 c - 20

3 x + 2 < 10