extreme f(x)=-(x/3)^3+2x^2+12x+1

Lösung

extrempunkte

f (x) = - (\frac{x}{3})^{3} + 2 x^{2} + 12 x + 1

Minimum (- 6 (23 - 3), 649 - 3843), Maximum (6 (3 + 23), 649 + 3843)

Schritte zur Lösung

f^{'} (x) = - \frac{x ^{2}}{9} + 4 x + 12

Intervalle finden:

Absteigend

: - \infty < x < - 123 + 18,

Ansteigend

: - 123 + 18 < x < 123 + 18,

Absteigend

: 123 + 18 < x < \infty

Stecke

x = - 6 (23 - 3)

- (\frac{x}{3})^{3} + 2 x^{2} + 12 x + 1 : 649 - 3843

Minimum (- 6 (23 - 3), 649 - 3843)

Stecke

x = 6 (3 + 23)

- (\frac{x}{3})^{3} + 2 x^{2} + 12 x + 1 : 649 + 3843

Maximum (6 (3 + 23), 649 + 3843)

Minimum (- 6 (23 - 3), 649 - 3843), Maximum (6 (3 + 23), 649 + 3843)

e x t re m e y = 5 + 6 x - 8 x^{3}

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

e x t re m e f (x) = (6 x + 1) e^{8 x - x^{2}}

e x t re m e f (x) = x^{4} - 3 x^{2}, - 2 \leq x \leq 0

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