extreme f(x)=x^3-x^2-8x+12

Lösung

extrempunkte

f (x) = x^{3} - x^{2} - 8 x + 12

Maximum (- \frac{4}{3}, \frac{500}{27}), Minimum (2, 0)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 2 x - 8

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{4}{3},

Absteigend

: - \frac{4}{3} < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - \frac{4}{3}

x^{3} - x^{2} - 8 x + 12 : \frac{500}{27}

Maximum (- \frac{4}{3}, \frac{500}{27})

Stecke

x = 2

x^{3} - x^{2} - 8 x + 12 : 0

Minimum (2, 0)

Maximum (- \frac{4}{3}, \frac{500}{27}), Minimum (2, 0)

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