monotone f(x)=x^3-4x^2+x+6

Lösung

monotone intervalle

f (x) = x^{3} - 4 x^{2} + x + 6

Ansteigend : - \infty < x < \frac{- 13 + 4}{3}, Absteigend : \frac{- 13 + 4}{3} < x < \frac{13 + 4}{3}, Ansteigend : \frac{13 + 4}{3} < x < \infty

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 8 x + 1

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 13 + 4}{3},

Absteigend

: \frac{- 13 + 4}{3} < x < \frac{13 + 4}{3},

Ansteigend

: \frac{13 + 4}{3} < x < \infty

Ansteigend : - \infty < x < \frac{- 13 + 4}{3}, Absteigend : \frac{- 13 + 4}{3} < x < \frac{13 + 4}{3}, Ansteigend : \frac{13 + 4}{3} < x < \infty

e x t re m e f (x) = 4 x^{2} - 24 x - 30

r an g e f (x) = - 3 y + 7 x = - 3

a sy m pt o t es f (x) = \frac{- 5 x ^{2} + 6 x - 2}{x ^{2} + 3}

s l o p e 9 x + y = 0

in t erce pt s f (x) = 5 x - 3 y = 15