extreme f(x)=2x^2-4x^4

Lösung

extrempunkte

f (x) = 2 x^{2} - 4 x^{4}

Maximum (- \frac{1}{2}, \frac{1}{4}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4})

Schritte zur Lösung

f^{'} (x) = 4 x - 16 x^{3}

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{1}{2},

Absteigend

: - \frac{1}{2} < x < 0,

Ansteigend

: 0 < x < \frac{1}{2},

Absteigend

: \frac{1}{2} < x < \infty

Stecke

x = - \frac{1}{2}

2 x^{2} - 4 x^{4} : \frac{1}{4}

Maximum (- \frac{1}{2}, \frac{1}{4})

Stecke

x = 0

2 x^{2} - 4 x^{4} : 0

Minimum (0, 0)

Stecke

x = \frac{1}{2}

2 x^{2} - 4 x^{4} : \frac{1}{4}

Maximum (\frac{1}{2}, \frac{1}{4})

Maximum (- \frac{1}{2}, \frac{1}{4}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4})

e x t re m e x^{3} + 6 x^{2} - 63 x + 10

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

e x t re m e f (x) = x^{2} + 10 x - 21

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

e x t re m e f (x, y) = \frac{x y + x - y + 2}{x ^{2} y + 1}