extreme f(x)=(x-5)e^{-x+5-(x-5)^2}

Lösung

extrempunkte

f (x) = (x - 5) e^{- x + 5 - (x - 5)^{2}}

Minimum (4, - 1), Maximum (\frac{11}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Schritte zur Lösung

f^{'} (x) = - e^{- (x - 4) (x - 5)} (x - 4) (2 x - 11)

Intervalle finden:

Absteigend

: - \infty < x < 4,

Ansteigend

: 4 < x < \frac{11}{2},

Absteigend

: \frac{11}{2} < x < \infty

Stecke

x = 4

(x - 5) e^{- x + 5 - (x - 5)^{2}} : - 1

Minimum (4, - 1)

Stecke

x = \frac{11}{2}

(x - 5) e^{- x + 5 - (x - 5)^{2}} : \frac{1}{2 e ^{\frac{3}{4}}}

Maximum (\frac{11}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

Minimum (4, - 1), Maximum (\frac{11}{2}, \frac{1}{2 e ^{\frac{3}{4}}})

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

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

e x t re m e f (x, y) = x y (1 + 2 x - 3 y)

e x t re m e f (x) = x^{2} - 4 x + 1, 0 \leq x \leq 3

e x t re m e x - 3