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

Lösung

extrempunkte

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

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

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

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

Stecke

x = - \frac{1}{3}

x^{3} - x + 1 : \frac{2}{3 3} + 1

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

Stecke

x = \frac{1}{3}

x^{3} - x + 1 : - \frac{2}{3 3} + 1

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

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

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

e x t re m e f (x) = \frac{1}{x}, 1 < x < 3

e x t re m e f (x, y) = x^{2} - y^{2} + 6 x y + 4 x - 8 y + 2

e x t re m e f (x) = \frac{2 x - 4}{x - 1}

e x t re m e y = x^{2} - 4 x - 5