extreme f(x)=x(6-2x)^2

Lösung

extrempunkte

f (x) = x (6 - 2 x)^{2}

Maximum (1, 16), Minimum (3, 0)

Schritte zur Lösung

f^{'} (x) = 12 (x - 3) (x - 1)

Intervalle finden:

Ansteigend

: - \infty < x < 1,

Absteigend

: 1 < x < 3,

Ansteigend

: 3 < x < \infty

Stecke

x = 1

x (6 - 2 x)^{2} : 16

Maximum (1, 16)

Stecke

x = 3

x (6 - 2 x)^{2} : 0

Minimum (3, 0)

Maximum (1, 16), Minimum (3, 0)

e x t re m e f (x) = \frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + \frac{2}{3}

e x t re m e f (x, y) = - 4 \cdot \frac{x ^{2} + 6 x + 8}{y ^{2} + 6 y + 6}

e x t re m e f (x) = x + \frac{108}{x ^{2}}

e x t re m e f (x) = \frac{3 x ^{2}}{\frac{1}{2} x ^{2} + 1}

e x t re m e g (x, y) = x y^{\frac{1}{2}}