extreme f(x)=4x^3-60x^2+200x

Lösung

extrempunkte

f (x) = 4 x^{3} - 60 x^{2} + 200 x

Maximum (\frac{5 ( 3 - 3 )}{3}, - 1000 (2 - 3) + \frac{18000 - 8000 3}{9}), Minimum (\frac{5 ( 3 + 3 )}{3}, - 1000 (2 + 3) + \frac{18000 + 8000 3}{9})

Schritte zur Lösung

f^{'} (x) = 12 x^{2} - 120 x + 200

Intervalle finden:

Ansteigend

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

Absteigend

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

Ansteigend

: \frac{5}{3} + 5 < x < \infty

Stecke

x = \frac{5 ( 3 - 3 )}{3}

4 x^{3} - 60 x^{2} + 200 x : - 1000 (2 - 3) + \frac{18000 - 8000 3}{9}

Maximum (\frac{5 ( 3 - 3 )}{3}, - 1000 (2 - 3) + \frac{18000 - 8000 3}{9})

Stecke

x = \frac{5 ( 3 + 3 )}{3}

4 x^{3} - 60 x^{2} + 200 x : - 1000 (2 + 3) + \frac{18000 + 8000 3}{9}

Minimum (\frac{5 ( 3 + 3 )}{3}, - 1000 (2 + 3) + \frac{18000 + 8000 3}{9})

Maximum (\frac{5 ( 3 - 3 )}{3}, - 1000 (2 - 3) + \frac{18000 - 8000 3}{9}), Minimum (\frac{5 ( 3 + 3 )}{3}, - 1000 (2 + 3) + \frac{18000 + 8000 3}{9})

e x t re m e f (x) = x^{3} + 2 y^{2} + 3 x y + 8

e x t re m e f (x, y) = x^{2} + x y + y^{2} - 6 x - 9 y

e x t re m e \frac{x + 3}{x + 2}

e x t re m e f (x) = x (\frac{39900 - x}{5})

e x t re m e f (x) = x^{3} - x^{2} - x + 5, - 1 \leq x \leq 2