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

Lösung

extrempunkte

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

Maximum (- 1, 7), Minimum (\frac{2}{3}, \frac{64}{27})

Schritte zur Lösung

f^{'} (x) = 6 x^{2} + 2 x - 4

Intervalle finden:

Ansteigend

: - \infty < x < - 1,

Absteigend

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

Ansteigend

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

Stecke

x = - 1

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

Maximum (- 1, 7)

Stecke

x = \frac{2}{3}

2 x^{3} + x^{2} - 4 x + 4 : \frac{64}{27}

Minimum (\frac{2}{3}, \frac{64}{27})

Maximum (- 1, 7), Minimum (\frac{2}{3}, \frac{64}{27})

e x t re m e f (x, y) = y (e^{x} - 1)

e x t re m e f (x) = \frac{x}{x ^{2} + 9}, - 4 \leq x \leq 5

e x t re m e x^{3} + 3 x^{2} - 24 x

e x t re m e - x^{3} + 4 x y - 2 y^{2} + 1

e x t re m e f (x) = - x^{3} + 12 x, (- 6, 5)