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

Lösung

extrempunkte

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

Minimum (- \frac{2 + 10}{2}, - \frac{e ^{2 + 10} ( 10 - 1 )}{2}), Maximum (\frac{- 2 + 10}{2}, \frac{e ^{2 - 10} ( 1 + 10 )}{2})

Schritte zur Lösung

f^{'} (x) = - e^{- 2 x} (2 x^{2} + 4 x - 3)

Intervalle finden:

Absteigend

: - \infty < x < - \frac{2 + 10}{2},

Ansteigend

: - \frac{2 + 10}{2} < x < \frac{5}{2} - 1,

Absteigend

: \frac{5}{2} - 1 < x < \infty

Stecke

x = - \frac{2 + 10}{2}

x (x + 3) e^{- 2 x} : - \frac{e ^{2 + 10} ( 10 - 1 )}{2}

Minimum (- \frac{2 + 10}{2}, - \frac{e ^{2 + 10} ( 10 - 1 )}{2})

Stecke

x = \frac{- 2 + 10}{2}

x (x + 3) e^{- 2 x} : \frac{e ^{2 - 10} ( 1 + 10 )}{2}

Maximum (\frac{- 2 + 10}{2}, \frac{e ^{2 - 10} ( 1 + 10 )}{2})

Minimum (- \frac{2 + 10}{2}, - \frac{e ^{2 + 10} ( 10 - 1 )}{2}), Maximum (\frac{- 2 + 10}{2}, \frac{e ^{2 - 10} ( 1 + 10 )}{2})

e x t re m e f (x) = - 45 x^{2} + 400 x

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

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

e x t re m e f (x) = x^{3} + 8 x + 12

e x t re m e 7 - 6 x - x^{3}