extreme f(x)=(x^3)/3-5x^2-50x+1

Lösung

extrempunkte

f (x) = \frac{x ^{3}}{3} - 5 x^{2} - 50 x + 1

Maximum (5 (1 - 3), \frac{125 ( 10 - 6 3 )}{3} + 5003 - 749), Minimum (5 (1 + 3), \frac{125 ( 10 + 6 3 )}{3} - 749 - 5003)

Schritte zur Lösung

f^{'} (x) = x^{2} - 10 x - 50

Intervalle finden:

Ansteigend

: - \infty < x < - 53 + 5,

Absteigend

: - 53 + 5 < x < 53 + 5,

Ansteigend

: 53 + 5 < x < \infty

Stecke

x = 5 (1 - 3)

\frac{x ^{3}}{3} - 5 x^{2} - 50 x + 1 : \frac{125 ( 10 - 6 3 )}{3} + 5003 - 749

Maximum (5 (1 - 3), \frac{125 ( 10 - 6 3 )}{3} + 5003 - 749)

Stecke

x = 5 (1 + 3)

\frac{x ^{3}}{3} - 5 x^{2} - 50 x + 1 : \frac{125 ( 10 + 6 3 )}{3} - 749 - 5003

Minimum (5 (1 + 3), \frac{125 ( 10 + 6 3 )}{3} - 749 - 5003)

Maximum (5 (1 - 3), \frac{125 ( 10 - 6 3 )}{3} + 5003 - 749), Minimum (5 (1 + 3), \frac{125 ( 10 + 6 3 )}{3} - 749 - 5003)

e x t re m e \frac{4}{x} + x^{4}

e x t re m e f (x, y) = 9 x^{2} - 4 y^{2}

e x t re m e f (x) = 120 x - x^{2} - (950 + 14 x)

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

e x t re m e f (x) = \frac{x ^{3}}{3} - 3 x^{2} + 8 x - 4