extreme f(x)=3x^3-4x^2-6x+1

Lösung

extrempunkte

f (x) = 3 x^{3} - 4 x^{2} - 6 x + 1

Maximum (\frac{4 - 70}{9}, \frac{140 70 - 776}{243} + 1), Minimum (\frac{4 + 70}{9}, \frac{- 776 - 140 70}{243} + 1)

Schritte zur Lösung

f^{'} (x) = 9 x^{2} - 8 x - 6

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 70 + 4}{9},

Absteigend

: \frac{- 70 + 4}{9} < x < \frac{70 + 4}{9},

Ansteigend

: \frac{70 + 4}{9} < x < \infty

Stecke

x = \frac{4 - 70}{9}

3 x^{3} - 4 x^{2} - 6 x + 1 : \frac{140 70 - 776}{243} + 1

Maximum (\frac{4 - 70}{9}, \frac{140 70 - 776}{243} + 1)

Stecke

x = \frac{4 + 70}{9}

3 x^{3} - 4 x^{2} - 6 x + 1 : \frac{- 776 - 140 70}{243} + 1

Minimum (\frac{4 + 70}{9}, \frac{- 776 - 140 70}{243} + 1)

Maximum (\frac{4 - 70}{9}, \frac{140 70 - 776}{243} + 1), Minimum (\frac{4 + 70}{9}, \frac{- 776 - 140 70}{243} + 1)

e x t re m e f (x, y) = - 6 x^{2} - y^{2} + 9 x - 5 y + 8

e x t re m e f (x) = x^{2} + \frac{240}{x}

e x t re m e f (x) = x + \frac{13}{x}

e x t re m e f (x) = \frac{2 x}{x ^{2} + 9}

e x t re m e f (x) = 3 x^{\frac{2}{3}} - 2 x [- 1.1]