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Beliebt Algebra >

4x^3-17x^2+21x-8>= 0

Lösung

4 x^{3} - 17 x^{2} + 21 x - 8 \geq 0

Lösung

\frac{13 - 41}{8} \leq x \leq 1 or x \geq \frac{13 + 41}{8}

+2

Intervall-Notation

[\frac{13 - 41}{8}, 1] \cup [\frac{13 + 41}{8}, \infty)

Dezimale

0.82460 \dots \leq x \leq 1 or x \geq 2.42539 \dots

Schritte zur Lösung

4 x^{3} - 17 x^{2} + 21 x - 8 \geq 0

Faktor

4 x^{3} - 17 x^{2} + 21 x - 8 : (x - 1) (x - \frac{13 + 41}{8}) (x - \frac{13 - 41}{8})

(x - 1) (x - \frac{13 + 41}{8}) (x - \frac{13 - 41}{8}) \geq 0

Identifiziere die Intervalle

\frac{13 - 41}{8} \leq x \leq 1 or x \geq \frac{13 + 41}{8}

Graph

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