extreme f(x)=-1+4x-x^3

Lösung

extrempunkte

f (x) = - 1 + 4 x - x^{3}

Minimum (- \frac{2}{3}, - \frac{16}{3 3} - 1), Maximum (\frac{2}{3}, \frac{16}{3 3} - 1)

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}

- 1 + 4 x - x^{3} : - \frac{16}{3 3} - 1

Minimum (- \frac{2}{3}, - \frac{16}{3 3} - 1)

Stecke

x = \frac{2}{3}

- 1 + 4 x - x^{3} : \frac{16}{3 3} - 1

Maximum (\frac{2}{3}, \frac{16}{3 3} - 1)

Minimum (- \frac{2}{3}, - \frac{16}{3 3} - 1), Maximum (\frac{2}{3}, \frac{16}{3 3} - 1)

in t erce pt s y = x^{2} - 3 x + 5

in v erse f (x) = - 2 x^{2} + 4 x - 1

in v erse f (x) = \frac{1}{3} (x + 1)^{2} - 2

in v erse f (x) = \frac{x ^{2}}{16}

e x t re m e f (x) = x^{3} - 2 x^{2} - 15 x + 10