extreme f(x)=x(27-41+2x)(41/2-x)

Lösung

extrempunkte

f (x) = x (27 - 41 + 2 x) (\frac{41}{2} - x)

Minimum (\frac{55 - 1303}{6}, \frac{24310 - 1303 1303}{54}), Maximum (\frac{55 + 1303}{6}, \frac{1303 1303 + 24310}{54})

Schritte zur Lösung

f^{'} (x) = - 6 x^{2} + 110 x - 287

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 1303 + 55}{6},

Ansteigend

: \frac{- 1303 + 55}{6} < x < \frac{1303 + 55}{6},

Absteigend

: \frac{1303 + 55}{6} < x < \infty

Stecke

x = \frac{55 - 1303}{6}

x (27 - 41 + 2 x) (\frac{41}{2} - x) : \frac{24310 - 1303 1303}{54}

Minimum (\frac{55 - 1303}{6}, \frac{24310 - 1303 1303}{54})

Stecke

x = \frac{55 + 1303}{6}

x (27 - 41 + 2 x) (\frac{41}{2} - x) : \frac{1303 1303 + 24310}{54}

Maximum (\frac{55 + 1303}{6}, \frac{1303 1303 + 24310}{54})

Minimum (\frac{55 - 1303}{6}, \frac{24310 - 1303 1303}{54}), Maximum (\frac{55 + 1303}{6}, \frac{1303 1303 + 24310}{54})

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

e x t re m e f (x) = (x + 1)^{7} - 7 x - 3

e x t re m e f (x) = - x^{2} + 4 x y - 5 y^{2} - 6 x + 22 y + 9

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

e x t re m e f (x, y) = x^{4}