extreme f(x)=-x(x+2)(x-2)

Lösung

extrempunkte

f (x) = - x (x + 2) (x - 2)

Minimum (- \frac{2}{3}, - \frac{16}{3 3}), Maximum (\frac{2}{3}, \frac{16}{3 3})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < - \frac{2}{3},

Ansteigend

: - \frac{2}{3} < x < \frac{2}{3},

Absteigend

: \frac{2}{3} < x < \infty

Stecke

x = - \frac{2}{3}

- x (x + 2) (x - 2) : - \frac{16}{3 3}

Minimum (- \frac{2}{3}, - \frac{16}{3 3})

Stecke

x = \frac{2}{3}

- x (x + 2) (x - 2) : \frac{16}{3 3}

Maximum (\frac{2}{3}, \frac{16}{3 3})

Minimum (- \frac{2}{3}, - \frac{16}{3 3}), Maximum (\frac{2}{3}, \frac{16}{3 3})

e x t re m e f (x) = 6 x^{5} - 10 x^{3}

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

e x t re m e 4 x^{3} + 3 x^{2} - 6 x + 1

e x t re m e f (x) = 5 x^{4} + 5 x^{3} + 7

e x t re m e f (x) = \frac{2}{5} x^{5} + 5 x^{4} + 16 x^{3} - 15