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

Lösung

extrempunkte

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

Maximum (0, - 5), Minimum (\frac{1}{3}, - \frac{137}{27})

Schritte zur Lösung

f^{'} (x) = 12 x^{2} - 4 x

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Absteigend

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

Ansteigend

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

Stecke

x = 0

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

Maximum (0, - 5)

Stecke

x = \frac{1}{3}

4 x^{3} - 2 x^{2} - 5 : - \frac{137}{27}

Minimum (\frac{1}{3}, - \frac{137}{27})

Maximum (0, - 5), Minimum (\frac{1}{3}, - \frac{137}{27})

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

e x t re m e f (x) = 5000 - 15 x + 0.08 x^{2}

e x t re m e 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

e x t re m e f (x) = - 2 x^{2} + 200 x

e x t re m e y = x^{2} - 5 x - 14