extreme y=x^4+2x^3-2x^2+1

Lösung

extrempunkte

y = x^{4} + 2 x^{3} - 2 x^{2} + 1

Minimum (- 2, - 7), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{13}{16})

Schritte zur Lösung

f^{'} (x) = 4 x^{3} + 6 x^{2} - 4 x

Intervalle finden:

Absteigend

: - \infty < x < - 2,

Ansteigend

: - 2 < x < 0,

Absteigend

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

Ansteigend

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

Stecke

x = - 2

x^{4} + 2 x^{3} - 2 x^{2} + 1 : - 7

Minimum (- 2, - 7)

Stecke

x = 0

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

Maximum (0, 1)

Stecke

x = \frac{1}{2}

x^{4} + 2 x^{3} - 2 x^{2} + 1 : \frac{13}{16}

Minimum (\frac{1}{2}, \frac{13}{16})

Minimum (- 2, - 7), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{13}{16})

l in e m = 16, (- 1, - \frac{9}{2})

d o main \frac{x ^{2} + 9 x + 20}{x + 4}

e x t re m e y = x + \frac{3600}{x}

d o main \frac{\frac{4}{x}}{\frac{4}{x} + 4}

in t erce pt s f (x) = \frac{3}{( x - 1 ) ( x ^{2} - 4 )}