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

Lösung

extrempunkte

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

Maximum (1, \frac{7}{3}), Minimum (5, - \frac{25}{3})

Schritte zur Lösung

f^{'} (x) = x^{2} - 6 x + 5

Intervalle finden:

Ansteigend

: - \infty < x < 1,

Absteigend

: 1 < x < 5,

Ansteigend

: 5 < x < \infty

Stecke

x = 1

\frac{1}{3} x^{3} - 3 x^{2} + 5 x : \frac{7}{3}

Maximum (1, \frac{7}{3})

Stecke

x = 5

\frac{1}{3} x^{3} - 3 x^{2} + 5 x : - \frac{25}{3}

Minimum (5, - \frac{25}{3})

Maximum (1, \frac{7}{3}), Minimum (5, - \frac{25}{3})

e x t re m e f (u, v) = 3 u - 7 v

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

e x t re m e sin (3 X)

e x t re m e f (x) = 2 x^{3} + 3 x^{2} - 72 x + 4

e x t re m e f (x) = x + cos (2 x), 0 \leq x \leq π