extreme f(x)=x^3+2x^2-4x+7

Lösung

extrempunkte

f (x) = x^{3} + 2 x^{2} - 4 x + 7

Maximum (- 2, 15), Minimum (\frac{2}{3}, \frac{149}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

: - 2 < x < \frac{2}{3},

Ansteigend

: \frac{2}{3} < x < \infty

Stecke

x = - 2

x^{3} + 2 x^{2} - 4 x + 7 : 15

Maximum (- 2, 15)

Stecke

x = \frac{2}{3}

x^{3} + 2 x^{2} - 4 x + 7 : \frac{149}{27}

Minimum (\frac{2}{3}, \frac{149}{27})

Maximum (- 2, 15), Minimum (\frac{2}{3}, \frac{149}{27})

e x t re m e f (x, y) = - 4 x y + 2 x^{2} - 10 y^{2} - 4 y^{3} - 1

e x t re m e 4 sin (2 x), - 2 π \leq x \leq π

e x t re m e 0.3 x^{2} - 66 x + 15239

e x t re m e a (- 6 - 0)^{2}

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