extreme f(x)= 1/3 x^3-4x^2+12x-5

Lösung

extrempunkte

f (x) = \frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5

Maximum (2, \frac{17}{3}), Minimum (6, - 5)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < 2,

Absteigend

: 2 < x < 6,

Ansteigend

: 6 < x < \infty

Stecke

x = 2

\frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5 : \frac{17}{3}

Maximum (2, \frac{17}{3})

Stecke

x = 6

\frac{1}{3} x^{3} - 4 x^{2} + 12 x - 5 : - 5

Minimum (6, - 5)

Maximum (2, \frac{17}{3}), Minimum (6, - 5)

e x t re m e f (x) = x 2 - x

e x t re m e f (x, y, z) = x + y 2 + x 3

e x t re m e h (x) = - 3 x^{5} + 5 x^{3}

e x t re m e f (x) = ln (5 - 2 x^{2})

e x t re m e f (x) = 2 x^{3} - \frac{11}{2} x^{2} - 10 x + 2