What is |x^3-7x+6|>= 19x-x^3-18 ?

The solution to |x^3-7x+6|>= 19x-x^3-18 is x = 3

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|x^3-7x+6|>= 19x-x^3-18

x^{3} - 7 x + 6 \geq 19 x - x^{3} - 18

x \leq 1 or x \geq 3

Interval Notation

(- \infty, 1] \cup [3, \infty)

Solution steps

x^{3} - 7 x + 6 \geq 19 x - x^{3} - 18

Find positive and negative intervals

x < - 3, - 3 \leq x \leq 1, 1 < x < 2, x \geq 2

Solve the inequality for each interval

x < - 3 or - 3 \leq x \leq 1 or No Solution or x \geq 3

Merge Overlapping Intervals

x \leq 1 or x \geq 3

\frac{1}{( 2 x - 1 )} > 3

∣ 2 x - 1 ∣ < - ∣ x + 3 ∣ - 2

∣ x ∣ > (\frac{1}{n})

∣ x + 1 ∣ + ∣ x - 2 ∣ > ∣ x ∣

∣ x - 3 ∣ < ∣ x - 2 ∣

What is |x^3-7x+6|>= 19x-x^3-18 ?
The solution to |x^3-7x+6|>= 19x-x^3-18 is x<= 1\lor x>= 3