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

Lösung

extrempunkte

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

Maximum (- \frac{2}{3}, \frac{121}{27}), Minimum (2, - 5)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

: 2 < x < \infty

Stecke

x = - \frac{2}{3}

x^{3} - 2 x^{2} - 4 x + 3 : \frac{121}{27}

Maximum (- \frac{2}{3}, \frac{121}{27})

Stecke

x = 2

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

Minimum (2, - 5)

Maximum (- \frac{2}{3}, \frac{121}{27}), Minimum (2, - 5)

d o main f (x) = 4 x - 12

a sy m pt o t es f (x) = \frac{( x + 2 )}{( x ^{2} - x - 6 )}

a sy m pt o t es \frac{3 x ^{2} + 1}{x ^{2} - 2 x - 3}

p a r i t y tan^{3} (x) d x

in t erce pt s 7 x^{2} - 40 x - 25