x^3-8x^2-x+8>= 0

Lösung

x^{3} - 8 x^{2} - x + 8 \geq 0

- 1 \leq x \leq 1 or x \geq 8

Intervall-Notation

[- 1, 1] \cup [8, \infty)

Schritte zur Lösung

x^{3} - 8 x^{2} - x + 8 \geq 0

Faktorisiere

x^{3} - 8 x^{2} - x + 8 : (x - 8) (x + 1) (x - 1)

(x - 8) (x + 1) (x - 1) \geq 0

Identifiziere die Intervalle

- 1 \leq x \leq 1 or x \geq 8

∣ 2 x - 2 ∣ < 0

s im pl i f y (4 - \frac{9}{2})^{2}

0 = 14 t - 1.86 t^{2}

- (- 2)^{2} + 3

- 2 y + 2 = - 7 y + 17