extreme f(x)=(x-3)e^{-x+3-(x-3)^2}

Lösung

extrempunkte

f (x) = (x - 3) e^{- x + 3 - (x - 3)^{2}}

Minimum (2, - 1), Maximum (\frac{7}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Schritte zur Lösung

f^{'} (x) = - e^{- (x - 2) (x - 3)} (x - 2) (2 x - 7)

Intervalle finden:

Absteigend

: - \infty < x < 2,

Ansteigend

: 2 < x < \frac{7}{2},

Absteigend

: \frac{7}{2} < x < \infty

Stecke

x = 2

(x - 3) e^{- x + 3 - (x - 3)^{2}} : - 1

Minimum (2, - 1)

Stecke

x = \frac{7}{2}

(x - 3) e^{- x + 3 - (x - 3)^{2}} : \frac{1}{2 e ^{\frac{3}{4}}}

Maximum (\frac{7}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Minimum (2, - 1), Maximum (\frac{7}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

e x t re m e f (x) = 2 x^{3} - 3 x^{2} - 12 x + 20

e x t re m e f (x) = (x - 9)^{3} (x + 5)^{2}

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

e x t re m e sin (4 x), 0 \leq x \leq \frac{π}{2}

e x t re m e f (x) = \frac{x ^{2} - 4 x + 4}{x - 6}