extreme f(x)=2x^3+4x^2-8x-11

Lösung

extrempunkte

f (x) = 2 x^{3} + 4 x^{2} - 8 x - 11

Maximum (- 2, 5), Minimum (\frac{2}{3}, - \frac{377}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < - 2,

Absteigend

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

Ansteigend

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

Stecke

x = - 2

2 x^{3} + 4 x^{2} - 8 x - 11 : 5

Maximum (- 2, 5)

Stecke

x = \frac{2}{3}

2 x^{3} + 4 x^{2} - 8 x - 11 : - \frac{377}{27}

Minimum (\frac{2}{3}, - \frac{377}{27})

Maximum (- 2, 5), Minimum (\frac{2}{3}, - \frac{377}{27})

d o main \frac{3}{4 \frac{5 + 3 x}{2}}

s hi f t 5 sin (6 x - π)

d o main \frac{2}{( x + 1 ) ^{3}}

d i s t an ce (3, 10), (- 2, - 2)

p a r i t y \frac{cos ( x )}{( sin ( x ) ) ^{0.5}}