extreme f(x)=x^4-x^2+1

Lösung

extrempunkte

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

Minimum (- \frac{1}{2}, \frac{3}{4}), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{3}{4})

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Absteigend

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

Ansteigend

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

Stecke

x = - \frac{1}{2}

x^{4} - x^{2} + 1 : \frac{3}{4}

Minimum (- \frac{1}{2}, \frac{3}{4})

Stecke

x = 0

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

Maximum (0, 1)

Stecke

x = \frac{1}{2}

x^{4} - x^{2} + 1 : \frac{3}{4}

Minimum (\frac{1}{2}, \frac{3}{4})

Minimum (- \frac{1}{2}, \frac{3}{4}), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{3}{4})

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

e x t re m e f (x) = x - x

e x t re m e f (x) = \frac{1}{16 - x ^{2} - y ^{2}}

e x t re m e f (x, y) = x + \frac{4}{x ^{2} y} + y

e x t re m e f (x, y) = - (x - y)^{4}