extreme f(χ)=χ^2e^{-4χ^2}

Lösung

extrempunkte

f (χ) = χ^{2} e^{- 4 χ^{2}}

Maximum (- \frac{1}{2}, \frac{1}{4 e}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4 e})

Schritte zur Lösung

f^{'} (χ) = - 2 e^{- 4 χ^{2}} χ (2 χ + 1) (2 χ - 1)

Intervalle finden:

Ansteigend

: - \infty < χ < - \frac{1}{2},

Absteigend

: - \frac{1}{2} < χ < 0,

Ansteigend

: 0 < χ < \frac{1}{2},

Absteigend

: \frac{1}{2} < χ < \infty

Stecke

χ = - \frac{1}{2}

χ^{2} e^{- 4 χ^{2}} : \frac{1}{4 e}

Maximum (- \frac{1}{2}, \frac{1}{4 e})

Stecke

χ = 0

χ^{2} e^{- 4 χ^{2}} : 0

Minimum (0, 0)

Stecke

χ = \frac{1}{2}

χ^{2} e^{- 4 χ^{2}} : \frac{1}{4 e}

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

Maximum (- \frac{1}{2}, \frac{1}{4 e}), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{4 e})

e x t re m e f (x) = \frac{y - 800}{- 2}

e x t re m e (x - 1) e^{\frac{1}{x}}

e x t re m e f (x) = x^{4} + x^{2} (y - 2) + 3 (y - 1)^{2}

e x t re m e f (x, y) = 2 e^{x} ln (1 + y)

e x t re m e f (x) = 9 (x - 4)^{2} + 5