extreme f(x)=x^3+8x^2+16x+3

Lösung

extrempunkte

f (x) = x^{3} + 8 x^{2} + 16 x + 3

Maximum (- 4, 3), Minimum (- \frac{4}{3}, - \frac{175}{27})

Schritte zur Lösung

f^{'} (x) = 3 x^{2} + 16 x + 16

Intervalle finden:

Ansteigend

: - \infty < x < - 4,

Absteigend

: - 4 < x < - \frac{4}{3},

Ansteigend

: - \frac{4}{3} < x < \infty

Stecke

x = - 4

x^{3} + 8 x^{2} + 16 x + 3 : 3

Maximum (- 4, 3)

Stecke

x = - \frac{4}{3}

x^{3} + 8 x^{2} + 16 x + 3 : - \frac{175}{27}

Minimum (- \frac{4}{3}, - \frac{175}{27})

Maximum (- 4, 3), Minimum (- \frac{4}{3}, - \frac{175}{27})

l in e m = - 4, (2, 7)

e x t re m e f (x) = x^{3} - 3 x^{2} + 2

in v erse f (x) = x^{5} - 4

in t erce pt s f (x) = 2 (x - 4)^{2} - 6

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