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

Lösung

extrempunkte

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

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

Schritte zur Lösung

f^{'} (x) = 8 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}

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

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

Stecke

x = 0

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

Minimum (0, 0)

Stecke

x = \frac{1}{2}

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

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

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

e x t re m e f (x, y) = (x - 1)^{2} + y^{2} + 5

e x t re m e f (x) = \frac{2 e ^{x}}{x ^{5}}, [0, \infty]

e x t re m e f (x) = x - 2 sin (x), (0, 2 π)

e x t re m e f (x) = \frac{- 1}{5 x ^{2} + 25 x + 30}

e x t re m e f (x) = \frac{5 x}{x ^{2} - 1}