(12x(x^3-2))/((x^3+1)^3)<= 0

Lösung

\frac{12 x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

x < - 1 or 0 \leq x \leq 32

Intervall-Notation

(- \infty, - 1) \cup [0, 32]

Dezimale

x < - 1 or 0 \leq x \leq 1.25992 \dots

Schritte zur Lösung

\frac{12 x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

Rewrite in standard form

\frac{x ( x ^{3} - 2 )}{( x ^{3} + 1 ) ^{3}} \leq 0

Faktor

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

\frac{x ( x - 3 2 ) ( x ^{2} + 3 2 x + 2 ^{\frac{2}{3}} )}{( x + 1 ) ^{3} ( x ^{2} - x + 1 ) ^{3}} \leq 0

Identifiziere die Intervalle

x < - 1 or 0 \leq x \leq 32

f = 4 (3.9822) - (3.9822)^{2}

3 x^{2} - 42 = 0

f a c t or x^{2} - 16 + 64

- 3 - 2 x > 9

w + 16 < 4