extreme f(x)=340x^2-2040x^3

Lösung

extrempunkte

f (x) = 340 x^{2} - 2040 x^{3}

Minimum (0, 0), Maximum (\frac{1}{9}, \frac{340}{243})

Schritte zur Lösung

f^{'} (x) = 680 x - 6120 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{1}{9},

Absteigend

: \frac{1}{9} < x < \infty

Stecke

x = 0

340 x^{2} - 2040 x^{3} : 0

Minimum (0, 0)

Stecke

x = \frac{1}{9}

340 x^{2} - 2040 x^{3} : \frac{340}{243}

Maximum (\frac{1}{9}, \frac{340}{243})

Minimum (0, 0), Maximum (\frac{1}{9}, \frac{340}{243})

e x t re m e f (x) = x (x - 1)

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e x t re m e f (x, y) = x^{2} + 6 x + y^{3} - 6 y^{2} + 10