extreme f(x)=x^3e^{-4x}

Lösung

extrempunkte

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

Sattel (0, 0), Maximum (\frac{3}{4}, \frac{27}{64 e ^{3}})

Schritte zur Lösung

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

Intervalle finden:

Ansteigend

: - \infty < x < 0,

Ansteigend

: 0 < x < \frac{3}{4},

Absteigend

: \frac{3}{4} < x < \infty

Stecke

x = 0

x^{3} e^{- 4 x} : 0

Sattel (0, 0)

Stecke

x = \frac{3}{4}

x^{3} e^{- 4 x} : \frac{27}{64 e ^{3}}

Maximum (\frac{3}{4}, \frac{27}{64 e ^{3}})

Sattel (0, 0), Maximum (\frac{3}{4}, \frac{27}{64 e ^{3}})

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

e x t re m e f (x) = 7 x^{2} - 28 x + 3, - 5 \leq x \leq 4

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

e x t re m e 6 x^{2} - 24 x - 30

e x t re m e 4 x y + 2 y - 3 x^{2} + 3 x - 4 y^{2}