extreme f(x)=2x^4-x^2+5

Lösung

extrempunkte

f (x) = 2 x^{4} - x^{2} + 5

Minimum (- \frac{1}{2}, \frac{39}{8}), Maximum (0, 5), Minimum (\frac{1}{2}, \frac{39}{8})

Schritte zur Lösung

f^{'} (x) = 8 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}

2 x^{4} - x^{2} + 5 : \frac{39}{8}

Minimum (- \frac{1}{2}, \frac{39}{8})

Stecke

x = 0

2 x^{4} - x^{2} + 5 : 5

Maximum (0, 5)

Stecke

x = \frac{1}{2}

2 x^{4} - x^{2} + 5 : \frac{39}{8}

Minimum (\frac{1}{2}, \frac{39}{8})

Minimum (- \frac{1}{2}, \frac{39}{8}), Maximum (0, 5), Minimum (\frac{1}{2}, \frac{39}{8})

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

e x t re m e y = 6 x^{4} + 8 x^{3}

s im pl i f y πj z + 3 w

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

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