extreme f(x)=x(22-40+2x)(40/2-x)

Lösung

extrempunkte

f (x) = x (22 - 40 + 2 x) (\frac{40}{2} - x)

Minimum (\frac{29 - 301}{3}, \frac{3596 - 1204 301}{27}), Maximum (\frac{29 + 301}{3}, \frac{1204 301 + 3596}{27})

Schritte zur Lösung

f^{'} (x) = - 2 (3 x^{2} - 58 x + 180)

Intervalle finden:

Absteigend

: - \infty < x < \frac{- 301 + 29}{3},

Ansteigend

: \frac{- 301 + 29}{3} < x < \frac{301 + 29}{3},

Absteigend

: \frac{301 + 29}{3} < x < \infty

Stecke

x = \frac{29 - 301}{3}

x (22 - 40 + 2 x) (\frac{40}{2} - x) : \frac{3596 - 1204 301}{27}

Minimum (\frac{29 - 301}{3}, \frac{3596 - 1204 301}{27})

Stecke

x = \frac{29 + 301}{3}

x (22 - 40 + 2 x) (\frac{40}{2} - x) : \frac{1204 301 + 3596}{27}

Maximum (\frac{29 + 301}{3}, \frac{1204 301 + 3596}{27})

Minimum (\frac{29 - 301}{3}, \frac{3596 - 1204 301}{27}), Maximum (\frac{29 + 301}{3}, \frac{1204 301 + 3596}{27})

e x t re m e f (x, y) = (y - 8) \cdot x^{2} - y^{2}

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

e x t re m e e^{6 x} + e^{- x}

e x t re m e F (x, y) = 12 x^{2} + 5 x y - 17 x + 7 y - 2 y^{2} + 5

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