extreme f(x)=x^3+x^2

Lösung

extrempunkte

f (x) = x^{3} + x^{2}

Maximum (- \frac{2}{3}, \frac{4}{27}), Minimum (0, 0)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

: - \frac{2}{3} < x < 0,

Ansteigend

: 0 < x < \infty

Stecke

x = - \frac{2}{3}

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

Maximum (- \frac{2}{3}, \frac{4}{27})

Stecke

x = 0

x^{3} + x^{2} : 0

Minimum (0, 0)

Maximum (- \frac{2}{3}, \frac{4}{27}), Minimum (0, 0)

e x t re m e f (x) = 2 x^{5} + 5 x^{4} - 15

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

d i s t an ce (5, - 2), (- 7, 3)

a sy m pt o t es f (x) = \frac{24 x ^{2} + 6 x}{2 x + 1}

r an g e f (x) = x^{2} + 2 x - 15