extreme f(x)=x(18-40+2x)(40/2-x)

Lösung

extrempunkte

f (x) = x (18 - 40 + 2 x) (\frac{40}{2} - x)

Minimum (\frac{31 - 301}{3}, \frac{- 3596 - 1204 301}{27}), Maximum (\frac{31 + 301}{3}, \frac{1204 301 - 3596}{27})

Schritte zur Lösung

f^{'} (x) = - 2 (3 x^{2} - 62 x + 220)

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 301 + 31}{3},

Ansteigend

: \frac{- 301 + 31}{3} < x < \frac{301 + 31}{3},

Absteigend

: \frac{301 + 31}{3} < x < \infty

Stecke

x = \frac{31 - 301}{3}

x (18 - 40 + 2 x) (\frac{40}{2} - x) : \frac{- 3596 - 1204 301}{27}

Minimum (\frac{31 - 301}{3}, \frac{- 3596 - 1204 301}{27})

Stecke

x = \frac{31 + 301}{3}

x (18 - 40 + 2 x) (\frac{40}{2} - x) : \frac{1204 301 - 3596}{27}

Maximum (\frac{31 + 301}{3}, \frac{1204 301 - 3596}{27})

Minimum (\frac{31 - 301}{3}, \frac{- 3596 - 1204 301}{27}), Maximum (\frac{31 + 301}{3}, \frac{1204 301 - 3596}{27})

e x t re m e f (x) = 0.001 x^{2} + 3.8 x - 70

e x t re m e f (x) = x^{3} - 27 x + 51

e x t re m e f (x) = x^{3} - 27 x + 50

e x t re m e g (x) = x^{3} - 3 x + 2

e x t re m e f (x) = x^{3} - 3 x + 11