extreme f(x)=x^3+x^2-5x-4

Lösung

extrempunkte

f (x) = x^{3} + x^{2} - 5 x - 4

Maximum (- \frac{5}{3}, \frac{67}{27}), Minimum (1, - 7)

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 - 4 : \frac{67}{27}

Maximum (- \frac{5}{3}, \frac{67}{27})

Stecke

x = 1

x^{3} + x^{2} - 5 x - 4 : - 7

Minimum (1, - 7)

Maximum (- \frac{5}{3}, \frac{67}{27}), Minimum (1, - 7)

e x t re m e f (x) = x 36 - x^{2}, - 6 \leq x \leq 6

e x t re m e f (x) = 5 x y

e x t re m e f (x, y) = x^{3} + y^{3} - 192 x - 300 y + 2

e x t re m e f (x) = \frac{4 x ^{2}}{x ^{2} + 3}

e x t re m e y = 2 x^{3} - 3 x^{2} - 12 x + 1