extreme f(x)=f(x)=x^3-9x^2+15x+4

Lösung

extrempunkte

f (x) = f (x) = x^{3} - 9 x^{2} + 15 x + 4

Maximum (1, 11), Minimum (5, - 21)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 18 x + 15

Intervalle finden:

Ansteigend

: - \infty < x < 1,

Absteigend

: 1 < x < 5,

Ansteigend

: 5 < x < \infty

Stecke

x = 1

x^{3} - 9 x^{2} + 15 x + 4 : 11

Maximum (1, 11)

Stecke

x = 5

x^{3} - 9 x^{2} + 15 x + 4 : - 21

Minimum (5, - 21)

Maximum (1, 11), Minimum (5, - 21)

e x t re m e f (x) = \frac{x ^{2} + 81}{4 x}, 1 \leq x \leq 12

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

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

e x t re m e f (x) = x e^{- \frac{x ^{2}}{32}} [- 3.8]

e x t re m e U (X, Y) = 2 X + 5 Y