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

Lösung

extrempunkte

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

Maximum (- \frac{1 + 19}{3}, 2 (1 + 19) + \frac{2 - 16 19}{27}), Minimum (\frac{- 1 + 19}{3}, - 2 (19 - 1) + \frac{16 19 + 2}{27})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 19 - 1}{3},

Absteigend

: \frac{- 19 - 1}{3} < x < \frac{19 - 1}{3},

Ansteigend

: \frac{19 - 1}{3} < x < \infty

Stecke

x = - \frac{1 + 19}{3}

x^{3} + x^{2} - 6 x : 2 (1 + 19) + \frac{2 - 16 19}{27}

Maximum (- \frac{1 + 19}{3}, 2 (1 + 19) + \frac{2 - 16 19}{27})

Stecke

x = \frac{- 1 + 19}{3}

x^{3} + x^{2} - 6 x : - 2 (19 - 1) + \frac{16 19 + 2}{27}

Minimum (\frac{- 1 + 19}{3}, - 2 (19 - 1) + \frac{16 19 + 2}{27})

Maximum (- \frac{1 + 19}{3}, 2 (1 + 19) + \frac{2 - 16 19}{27}), Minimum (\frac{- 1 + 19}{3}, - 2 (19 - 1) + \frac{16 19 + 2}{27})

e x t re m e f (x, y) = x^{2} - 2 x + y^{2} + 1

e x t re m e \frac{x + 3}{x - 2}

e x t re m e \frac{3}{4} \frac{2}{5}

e x t re m e f (x) = 3 - 4 x - 5 x^{2} [- 2.4]

e x t re m e f (x) = \frac{x + 2}{( x + 1 ) ( 2 - x )}