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

Lösung

extrempunkte

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

Maximum (- \frac{7}{3}, \frac{392}{27}), Minimum (1, - 4)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{7}{3},

Absteigend

: - \frac{7}{3} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{7}{3}

x^{3} + 2 x^{2} - 7 x : \frac{392}{27}

Maximum (- \frac{7}{3}, \frac{392}{27})

Stecke

x = 1

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

Minimum (1, - 4)

Maximum (- \frac{7}{3}, \frac{392}{27}), Minimum (1, - 4)

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

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

e x t re m e \frac{1}{x - 2}

e x t re m e f (x, y) = x^{2} + y^{2} - 6 y + 9

e x t re m e f (x) = x^{2}, - 1 \leq x \leq 0