extreme f(x)=x(10-2x)(12-2x)

Lösung

extrempunkte

f (x) = x (10 - 2 x) (12 - 2 x)

Maximum (\frac{11 - 31}{3}, \frac{1232 + 248 31}{27}), Minimum (\frac{11 + 31}{3}, \frac{1232 - 248 31}{27})

Schritte zur Lösung

f^{'} (x) = 4 (3 x^{2} - 22 x + 30)

Intervalle finden:

Ansteigend

: - \infty < x < \frac{- 31 + 11}{3},

Absteigend

: \frac{- 31 + 11}{3} < x < \frac{31 + 11}{3},

Ansteigend

: \frac{31 + 11}{3} < x < \infty

Stecke

x = \frac{11 - 31}{3}

x (10 - 2 x) (12 - 2 x) : \frac{1232 + 248 31}{27}

Maximum (\frac{11 - 31}{3}, \frac{1232 + 248 31}{27})

Stecke

x = \frac{11 + 31}{3}

x (10 - 2 x) (12 - 2 x) : \frac{1232 - 248 31}{27}

Minimum (\frac{11 + 31}{3}, \frac{1232 - 248 31}{27})

Maximum (\frac{11 - 31}{3}, \frac{1232 + 248 31}{27}), Minimum (\frac{11 + 31}{3}, \frac{1232 - 248 31}{27})

e x t re m e f (x) = (x - 2) (x + 5) (x + 2)

e x t re m e f (x) = x^{3} - 75 x + 7

e x t re m e f (x) = x ln (x) - x

e x t re m e f (x) = 6 + 3 x - 3 x^{2}, 0 \leq x \leq 3

e x t re m e \frac{x ^{2} + x}{x - 1}