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

Lösung

extrempunkte

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

Maximum (\frac{4 - 19}{3}, \frac{38 19 - 164}{27} + 4), Minimum (\frac{4 + 19}{3}, \frac{- 164 - 38 19}{27} + 4)

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 19 + 4}{3},

Absteigend

: \frac{- 19 + 4}{3} < x < \frac{19 + 4}{3},

Ansteigend

: \frac{19 + 4}{3} < x < \infty

Stecke

x = \frac{4 - 19}{3}

x^{3} - 4 x^{2} - x + 4 : \frac{38 19 - 164}{27} + 4

Maximum (\frac{4 - 19}{3}, \frac{38 19 - 164}{27} + 4)

Stecke

x = \frac{4 + 19}{3}

x^{3} - 4 x^{2} - x + 4 : \frac{- 164 - 38 19}{27} + 4

Minimum (\frac{4 + 19}{3}, \frac{- 164 - 38 19}{27} + 4)

Maximum (\frac{4 - 19}{3}, \frac{38 19 - 164}{27} + 4), Minimum (\frac{4 + 19}{3}, \frac{- 164 - 38 19}{27} + 4)

in t erce pt s f (x) = x^{2} + 4 x + 3

mi d p o in t (4, - 1), (- 1, - 2)

p a r i t y f (x) = \frac{x ^{2}}{2 - x}

s im pl i f y (- 1.2) (7.3)

r an g e \frac{1}{2} x + 3