extreme f(x)=x^3-8x^2-12x+1

Lösung

extrempunkte

f (x) = x^{3} - 8 x^{2} - 12 x + 1

Maximum (- \frac{2}{3}, \frac{139}{27}), Minimum (6, - 143)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 16 x - 12

Intervalle finden:

Ansteigend

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

Absteigend

: - \frac{2}{3} < x < 6,

Ansteigend

: 6 < x < \infty

Stecke

x = - \frac{2}{3}

x^{3} - 8 x^{2} - 12 x + 1 : \frac{139}{27}

Maximum (- \frac{2}{3}, \frac{139}{27})

Stecke

x = 6

x^{3} - 8 x^{2} - 12 x + 1 : - 143

Minimum (6, - 143)

Maximum (- \frac{2}{3}, \frac{139}{27}), Minimum (6, - 143)

e x t re m e 4 x + 6 y - x^{2} - y^{2}

e x t re m e 2 x^{3} - 2 x^{2} - 2 x + 9

e x t re m e f (x, y) = x^{2} - 4 x + y^{2} - 6 y - 2

e x t re m e f (x) = 4 x^{3} - x^{2} + 9

e x t re m e x^{2} - 6 x + 12