extreme f(x)=x^3-2x^2-x+1

Lösung

extrempunkte

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

Maximum (\frac{2 - 7}{3}, \frac{14 7 - 34}{27} + 1), Minimum (\frac{2 + 7}{3}, \frac{- 34 - 14 7}{27} + 1)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} - 4 x - 1

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 7 + 2}{3},

Absteigend

: \frac{- 7 + 2}{3} < x < \frac{7 + 2}{3},

Ansteigend

: \frac{7 + 2}{3} < x < \infty

Stecke

x = \frac{2 - 7}{3}

x^{3} - 2 x^{2} - x + 1 : \frac{14 7 - 34}{27} + 1

Maximum (\frac{2 - 7}{3}, \frac{14 7 - 34}{27} + 1)

Stecke

x = \frac{2 + 7}{3}

x^{3} - 2 x^{2} - x + 1 : \frac{- 34 - 14 7}{27} + 1

Minimum (\frac{2 + 7}{3}, \frac{- 34 - 14 7}{27} + 1)

Maximum (\frac{2 - 7}{3}, \frac{14 7 - 34}{27} + 1), Minimum (\frac{2 + 7}{3}, \frac{- 34 - 14 7}{27} + 1)

in v erse f (x) = \frac{1 - x}{x + 2}

a sy m pt o t es f (x) = \frac{1}{x - 3} - 2

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

in v erse f (x) = 5 + e^{2 x + 4}

a sy m pt o t es f (x) = \frac{x ^{2} - 9}{x ( x - 3 )}