extreme f(x)=2x^3-42x^2+270x+10

Lösung

extrempunkte

f (x) = 2 x^{3} - 42 x^{2} + 270 x + 10

Maximum (5, 560), Minimum (9, 496)

Schritte zur Lösung

f^{'} (x) = 6 x^{2} - 84 x + 270

Intervalle finden:

Ansteigend

: - \infty < x < 5,

Absteigend

: 5 < x < 9,

Ansteigend

: 9 < x < \infty

Stecke

x = 5

2 x^{3} - 42 x^{2} + 270 x + 10 : 560

Maximum (5, 560)

Stecke

x = 9

2 x^{3} - 42 x^{2} + 270 x + 10 : 496

Minimum (9, 496)

Maximum (5, 560), Minimum (9, 496)

e x t re m e - x^{3} + 27 x - 55

e x t re m e f (x, y, λ) = 9 - x^{2} - y^{2} + λ (4 - x - y)

e x t re m e f (x, y) = 7 x^{2} + 6 x y + 5 y^{2} + 4 x y^{2} + 7 x^{2} y

e x t re m e f (x) = 8 x - 2 x^{2} + 12 x y - 12 y^{2} + y

e x t re m e f (x) = 2 x^{2} + 3 x - 5, - 4 \leq x \leq 4