extreme f(x)=(x-2)(x+2)(x+4)

Lösung

extrempunkte

f (x) = (x - 2) (x + 2) (x + 4)

Maximum (- \frac{2 ( 2 + 7 )}{3}, \frac{272 + 112 7}{27} - 16), Minimum (\frac{2 ( 7 - 2 )}{3}, \frac{272 - 112 7}{27} - 16)

Schritte zur Lösung

f^{'} (x) = 3 x^{2} + 8 x - 4

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 2 7 - 4}{3},

Absteigend

: \frac{- 2 7 - 4}{3} < x < \frac{2 7 - 4}{3},

Ansteigend

: \frac{2 7 - 4}{3} < x < \infty

Stecke

x = - \frac{2 ( 2 + 7 )}{3}

(x - 2) (x + 2) (x + 4) : \frac{272 + 112 7}{27} - 16

Maximum (- \frac{2 ( 2 + 7 )}{3}, \frac{272 + 112 7}{27} - 16)

Stecke

x = \frac{2 ( 7 - 2 )}{3}

(x - 2) (x + 2) (x + 4) : \frac{272 - 112 7}{27} - 16

Minimum (\frac{2 ( 7 - 2 )}{3}, \frac{272 - 112 7}{27} - 16)

Maximum (- \frac{2 ( 2 + 7 )}{3}, \frac{272 + 112 7}{27} - 16), Minimum (\frac{2 ( 7 - 2 )}{3}, \frac{272 - 112 7}{27} - 16)

e x t re m e f (x) = 11 + 5 x + x^{2}

e x t re m e f (x) = x^{2} lo g_{3} (x)

e x t re m e f (x) = \frac{x ^{2} - 9 x + 50}{x - 7}

e x t re m e - \frac{4}{x - 7}

e x t re m e f (x, y) = - 4 x^{2} - 2 y^{2} - 8 x + 12 y + 5