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

Lösung

extrempunkte

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

Maximum (- 3, 13), Minimum (1, \frac{7}{3})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - 3

\frac{x ^{3}}{3} + x^{2} - 3 x + 4 : 13

Maximum (- 3, 13)

Stecke

x = 1

\frac{x ^{3}}{3} + x^{2} - 3 x + 4 : \frac{7}{3}

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

Maximum (- 3, 13), Minimum (1, \frac{7}{3})

e x t re m e f (x) = x^{2} - 10 x - 1, 2 \leq x \leq 6

e x t re m e f (x) = 3 3 x^{3} + 24

e x t re m e 3 x (28 - x)

e x t re m e f (x) = x^{3} + 15 x - 9

e x t re m e f (x) = x^{3} - 2 x^{2} - 15 x + 52.2