extreme f(x)=x^6-x^4

Lösung

extrempunkte

f (x) = x^{6} - x^{4}

Minimum (- \frac{2}{3}, - \frac{4}{27}), Maximum (0, 0), Minimum (\frac{2}{3}, - \frac{4}{27})

Schritte zur Lösung

f^{'} (x) = 6 x^{5} - 4 x^{3}

Intervalle finden:

Absteigend

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

Ansteigend

: - \frac{2}{3} < x < 0,

Absteigend

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

Ansteigend

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

Stecke

x = - \frac{2}{3}

x^{6} - x^{4} : - \frac{4}{27}

Minimum (- \frac{2}{3}, - \frac{4}{27})

Stecke

x = 0

x^{6} - x^{4} : 0

Maximum (0, 0)

Stecke

x = \frac{2}{3}

x^{6} - x^{4} : - \frac{4}{27}

Minimum (\frac{2}{3}, - \frac{4}{27})

Minimum (- \frac{2}{3}, - \frac{4}{27}), Maximum (0, 0), Minimum (\frac{2}{3}, - \frac{4}{27})

e x t re m e f (x, y) = 3 x^{(4)} y^{(5)} + 6 x - 7 y

e x t re m e \frac{4 x}{x ^{2} + 4}

e x t re m e f (x) = - 0.018 x^{3} + 0.215 x^{2} + 0.057 x + 1

e x t re m e f (x) = 2 x^{3} + 9 x^{2} - 60 x + 7, - 5 \leq x \leq 5

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