extreme f(x)=2x^4-8x^2+6

Lösung

extrempunkte

f (x) = 2 x^{4} - 8 x^{2} + 6

Minimum (- 2, - 2), Maximum (0, 6), Minimum (2, - 2)

Schritte zur Lösung

f^{'} (x) = 8 x^{3} - 16 x

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < 0,

Absteigend

: 0 < x < 2,

Ansteigend

: 2 < x < \infty

Stecke

x = - 2

2 x^{4} - 8 x^{2} + 6 : - 2

Minimum (- 2, - 2)

Stecke

x = 0

2 x^{4} - 8 x^{2} + 6 : 6

Maximum (0, 6)

Stecke

x = 2

2 x^{4} - 8 x^{2} + 6 : - 2

Minimum (2, - 2)

Minimum (- 2, - 2), Maximum (0, 6), Minimum (2, - 2)

a sy m pt o t es f (x) = \frac{x ^{3} + 1}{x ^{2}}

in v erse y = \frac{1}{2} x - \frac{3}{2}

in v erse \frac{x - 2}{3 x - 4}

sy mm e t ry y = - 5 x^{2} + 20 x

d o main f (x) = 16 - x^{2} + x + 1