extreme f(x)=3x^3-9x^2-315x+7

Lösung

extrempunkte

f (x) = 3 x^{3} - 9 x^{2} - 315 x + 7

Maximum (- 5, 982), Minimum (7, - 1610)

Schritte zur Lösung

f^{'} (x) = 9 x^{2} - 18 x - 315

Intervalle finden:

Ansteigend

: - \infty < x < - 5,

Absteigend

: - 5 < x < 7,

Ansteigend

: 7 < x < \infty

Stecke

x = - 5

3 x^{3} - 9 x^{2} - 315 x + 7 : 982

Maximum (- 5, 982)

Stecke

x = 7

3 x^{3} - 9 x^{2} - 315 x + 7 : - 1610

Minimum (7, - 1610)

Maximum (- 5, 982), Minimum (7, - 1610)

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

e x t re m e 2 x^{3} + 6 x^{2} - 90 x + 7, - 5 \leq x \leq 6

e x t re m e f (x) = - \frac{5}{4} x^{2} + \frac{5}{2} x + \frac{7}{4}

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

e x t re m e - x^{2} + 8 x - 9