extreme f(x)=x(19-37+2x)(37/2-x)

Lösung

extrempunkte

f (x) = x (19 - 37 + 2 x) (\frac{37}{2} - x)

Minimum (\frac{55 - 1027}{6}, \frac{1540 - 1027 1027}{54}), Maximum (\frac{55 + 1027}{6}, \frac{1027 1027 + 1540}{54})

Schritte zur Lösung

f^{'} (x) = - 6 x^{2} + 110 x - 333

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 1027 + 55}{6},

Ansteigend

: \frac{- 1027 + 55}{6} < x < \frac{1027 + 55}{6},

Absteigend

: \frac{1027 + 55}{6} < x < \infty

Stecke

x = \frac{55 - 1027}{6}

x (19 - 37 + 2 x) (\frac{37}{2} - x) : \frac{1540 - 1027 1027}{54}

Minimum (\frac{55 - 1027}{6}, \frac{1540 - 1027 1027}{54})

Stecke

x = \frac{55 + 1027}{6}

x (19 - 37 + 2 x) (\frac{37}{2} - x) : \frac{1027 1027 + 1540}{54}

Maximum (\frac{55 + 1027}{6}, \frac{1027 1027 + 1540}{54})

Minimum (\frac{55 - 1027}{6}, \frac{1540 - 1027 1027}{54}), Maximum (\frac{55 + 1027}{6}, \frac{1027 1027 + 1540}{54})

e x t re m e y = x^{2} (x^{2} - 5)

e x t re m e f (x) = 6 x + 18

e x t re m e f (x) = \frac{4 y + 7}{3 - x}

e x t re m e 3 cos^{2} (x), 0 \leq x \leq π

e x t re m e f (x) = x + \frac{9}{x} + 2, 1 \leq x \leq 18