extreme f(x)=x^3+x^2-5x-5

Lösung

extrempunkte

f (x) = x^{3} + x^{2} - 5 x - 5

Maximum (- \frac{5}{3}, \frac{40}{27}), Minimum (1, - 8)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} + 2 x - 5

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{5}{3},

Absteigend

: - \frac{5}{3} < x < 1,

Ansteigend

: 1 < x < \infty

Stecke

x = - \frac{5}{3}

x^{3} + x^{2} - 5 x - 5 : \frac{40}{27}

Maximum (- \frac{5}{3}, \frac{40}{27})

Stecke

x = 1

x^{3} + x^{2} - 5 x - 5 : - 8

Minimum (1, - 8)

Maximum (- \frac{5}{3}, \frac{40}{27}), Minimum (1, - 8)

e x t re m e f (x, y) = 5 x^{2} + 3 x y + e^{x y}

e x t re m e f (x) = \frac{1}{9} \cdot (\frac{x ^{3}}{x + 2})

e x t re m e f (x) = x^{3} - 5 x^{2} + 3 x + 10, - 2 \leq x \leq 4

e x t re m e f (x, y) = x^{4} - 2 x^{2} + y^{3} - 3 y

e x t re m e ln (x^{4} + 27)