extreme f(x)=14x^2-x^3

Lösung

extrempunkte

f (x) = 14 x^{2} - x^{3}

Minimum (0, 0), Maximum (\frac{28}{3}, \frac{10976}{27})

Schritte zur Lösung

f^{'} (x) = 28 x - 3 x^{2}

Intervalle finden:

Absteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{28}{3},

Absteigend

: \frac{28}{3} < x < \infty

Stecke

x = 0

14 x^{2} - x^{3} : 0

Minimum (0, 0)

Stecke

x = \frac{28}{3}

14 x^{2} - x^{3} : \frac{10976}{27}

Maximum (\frac{28}{3}, \frac{10976}{27})

Minimum (0, 0), Maximum (\frac{28}{3}, \frac{10976}{27})

e x t re m e f (x) = x^{3} - 3 x^{2} + 6 x - 8

e x t re m e f (x) = \frac{4 ( x - 10 ) ^{2}}{x - 6}, (0, 6)

e x t re m e f (x) = 15 x^{\frac{2}{3}} - x

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

e x t re m e y = \frac{x ^{2} - 9 x + 39}{x - 7}