extreme f(x)=(x^3)/3-16x^2+240x-36

Lösung

extrempunkte

f (x) = \frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36

Maximum (12, 1116), Minimum (20, \frac{3092}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} - 32 x + 240

Intervalle finden:

Ansteigend

: - \infty < x < 12,

Absteigend

: 12 < x < 20,

Ansteigend

: 20 < x < \infty

Stecke

x = 12

\frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36 : 1116

Maximum (12, 1116)

Stecke

x = 20

\frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36 : \frac{3092}{3}

Minimum (20, \frac{3092}{3})

Maximum (12, 1116), Minimum (20, \frac{3092}{3})

e x t re m e y = 4 x^{3} - 8 x

e x t re m e f (x) = \frac{7}{4 - 3 sin ^{2} ( x )}

e x t re m e x^{2} - 5, 1 < x < 7

e x t re m e (x - 7) (x - 5)^{2} (x - 3) (x - 1)

e x t re m e \frac{400}{1 + 8 e ^{0.05 x}}