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

Lösung

extrempunkte

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

Maximum (- 2, \frac{13}{3}), Minimum (1, - \frac{1}{6})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - 2

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 1 : \frac{13}{3}

Maximum (- 2, \frac{13}{3})

Stecke

x = 1

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 2 x + 1 : - \frac{1}{6}

Minimum (1, - \frac{1}{6})

Maximum (- 2, \frac{13}{3}), Minimum (1, - \frac{1}{6})

e x t re m e f (x, y) = \frac{1}{3} x^{2} + 2 x y + 3 y^{2} + 4 x - 1

e x t re m e f (x, y) = 4 x + 6 y - x^{2} - y^{2} + 8

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

e x t re m e f (x) = \frac{4 x - 12}{( x - 2 ) ^{2}}

e x t re m e f (x) = \frac{I n x}{x}