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

Lösung

extrempunkte

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

Minimum (- \frac{1}{3}, - \frac{32}{27}), Maximum (1, 0)

Schritte zur Lösung

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

Intervalle finden:

Absteigend

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

Ansteigend

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

Absteigend

: 1 < x < \infty

Stecke

x = - \frac{1}{3}

- x^{3} + x^{2} + x - 1 : - \frac{32}{27}

Minimum (- \frac{1}{3}, - \frac{32}{27})

Stecke

x = 1

- x^{3} + x^{2} + x - 1 : 0

Maximum (1, 0)

Minimum (- \frac{1}{3}, - \frac{32}{27}), Maximum (1, 0)

e x t re m e f (x) = - x^{3} + x^{2} + x + 6

e x t re m e f (x) = x y - 2 x - y

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

e x t re m e f (x) = (x + 2)^{3} (x - 3)^{2}

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