extreme f(x)=x(25-46+2x)(46/2-x)

Lösung

extrempunkte

f (x) = x (25 - 46 + 2 x) (\frac{46}{2} - x)

Minimum (\frac{67 - 1591}{6}, \frac{9514 - 1591 1591}{54}), Maximum (\frac{67 + 1591}{6}, \frac{1591 1591 + 9514}{54})

Schritte zur Lösung

f^{'} (x) = - 6 x^{2} + 134 x - 483

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 1591 + 67}{6},

Ansteigend

: \frac{- 1591 + 67}{6} < x < \frac{1591 + 67}{6},

Absteigend

: \frac{1591 + 67}{6} < x < \infty

Stecke

x = \frac{67 - 1591}{6}

x (25 - 46 + 2 x) (\frac{46}{2} - x) : \frac{9514 - 1591 1591}{54}

Minimum (\frac{67 - 1591}{6}, \frac{9514 - 1591 1591}{54})

Stecke

x = \frac{67 + 1591}{6}

x (25 - 46 + 2 x) (\frac{46}{2} - x) : \frac{1591 1591 + 9514}{54}

Maximum (\frac{67 + 1591}{6}, \frac{1591 1591 + 9514}{54})

Minimum (\frac{67 - 1591}{6}, \frac{9514 - 1591 1591}{54}), Maximum (\frac{67 + 1591}{6}, \frac{1591 1591 + 9514}{54})

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

e x t re m e y = x^{5} - 15 x^{3} + 100

e x t re m e f (x) = (ln (x^{2} + 1))

e x t re m e f (x) = 3 x + 5 x^{- 1}

e x t re m e f (x) = (x - 1)^{5}