extreme f(x)=x^3-2x-1

Lösung

extrempunkte

f (x) = x^{3} - 2 x - 1

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} - 1), Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} - 1)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{2}{3},

Absteigend

: - \frac{2}{3} < x < \frac{2}{3},

Ansteigend

: \frac{2}{3} < x < \infty

Stecke

x = - \frac{2}{3}

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

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} - 1)

Stecke

x = \frac{2}{3}

x^{3} - 2 x - 1 : (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} - 1

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

Maximum (- \frac{2}{3}, - (\frac{2}{3})^{\frac{3}{2}} + 2 \frac{2}{3} - 1), Minimum (\frac{2}{3}, (\frac{2}{3})^{\frac{3}{2}} - 2 \frac{2}{3} - 1)

y = x^{2} + 4 x + 4

d o main cos (3 t)

cr i t i c a l f (x) = 3 - 3 x + 3 x^{2} - x^{3}

p er p e n d i c u l a r y = - \frac{1}{4} x + 3, (3, - 3)

p a r i t y f (x) = x^{2} (x - 2)^{3} (x + 3)