extreme f(x)=(x^3)/3+(x^2)/2-6x+10

Lösung

extrempunkte

f (x) = \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 10

Maximum (- 3, \frac{47}{2}), Minimum (2, \frac{8}{3})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 3,

Absteigend

: - 3 < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 3

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 10 : \frac{47}{2}

Maximum (- 3, \frac{47}{2})

Stecke

x = 2

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + 10 : \frac{8}{3}

Minimum (2, \frac{8}{3})

Maximum (- 3, \frac{47}{2}), Minimum (2, \frac{8}{3})

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

e x t re m e y = 3 x^{2} = 12 x + 11

e x t re m e f (x) - \frac{1}{2} (x + 1)^{2} - \frac{3}{2}

e x t re m e 2 x + \frac{2}{3}

e x t re m e f (x) = x^{8} (x - 3)^{7}