extreme f(x)=x(25-39+2x)(39/2-x)

Lösung

extrempunkte

f (x) = x (25 - 39 + 2 x) (\frac{39}{2} - x)

Minimum (\frac{53 - 1171}{6}, \frac{18656 - 1171 1171}{54}), Maximum (\frac{53 + 1171}{6}, \frac{1171 1171 + 18656}{54})

Schritte zur Lösung

f^{'} (x) = - 6 x^{2} + 106 x - 273

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 1171 + 53}{6},

Ansteigend

: \frac{- 1171 + 53}{6} < x < \frac{1171 + 53}{6},

Absteigend

: \frac{1171 + 53}{6} < x < \infty

Stecke

x = \frac{53 - 1171}{6}

x (25 - 39 + 2 x) (\frac{39}{2} - x) : \frac{18656 - 1171 1171}{54}

Minimum (\frac{53 - 1171}{6}, \frac{18656 - 1171 1171}{54})

Stecke

x = \frac{53 + 1171}{6}

x (25 - 39 + 2 x) (\frac{39}{2} - x) : \frac{1171 1171 + 18656}{54}

Maximum (\frac{53 + 1171}{6}, \frac{1171 1171 + 18656}{54})

Minimum (\frac{53 - 1171}{6}, \frac{18656 - 1171 1171}{54}), Maximum (\frac{53 + 1171}{6}, \frac{1171 1171 + 18656}{54})

e x t re m e f (x) = x^{2} + 8 x + 11

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

e x t re m e f (y) = x^{3} - 9 x^{2} + 7 x - 6

e x t re m e sin (3 x), - \frac{π}{4} \leq x \leq \frac{π}{3}

e x t re m e f (x) = 2 x^{3} - 39 x^{2} + 240 x - 9