extreme f(x)=4x^3-440x^2+10500x

Lösung

extrempunkte

f (x) = 4 x^{3} - 440 x^{2} + 10500 x

Maximum (15, 72000), Minimum (\frac{175}{3}, - \frac{2450000}{27})

Schritte zur Lösung

f^{'} (x) = 12 x^{2} - 880 x + 10500

Intervalle finden:

Ansteigend

: - \infty < x < 15,

Absteigend

: 15 < x < \frac{175}{3},

Ansteigend

: \frac{175}{3} < x < \infty

Stecke

x = 15

4 x^{3} - 440 x^{2} + 10500 x : 72000

Maximum (15, 72000)

Stecke

x = \frac{175}{3}

4 x^{3} - 440 x^{2} + 10500 x : - \frac{2450000}{27}

Minimum (\frac{175}{3}, - \frac{2450000}{27})

Maximum (15, 72000), Minimum (\frac{175}{3}, - \frac{2450000}{27})

e x t re m e f (x) = 4 + 15 x + 6 x^{2} - x^{3}

e x t re m e g (x, y) = 4 x + 9 y

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

e x t re m e f (x) = 2 x - 6 sin (x)

e x t re m e f (x, y) = 2 x^{3} - 6 x y + y^{2} + 500