extreme f(x)=(x^2-4)/(e^{x^4-1)}

Lösung

extrempunkte

f (x) = \frac{x ^{2} - 4}{e ^{x^{4} - 1}}

Maximum - \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}, Minimum (0, - 4 e), Maximum \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \frac{4 + 3 2}{2} < x < 0,

Ansteigend

: 0 < x < \frac{4 + 3 2}{2}

Stecke

x = - \frac{4 + 3 2}{2}

\frac{x ^{2} - 4}{e ^{x^{4} - 1}} : \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Maximum - \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Stecke

x = 0

\frac{x ^{2} - 4}{e ^{x^{4} - 1}} : - 4 e

Minimum (0, - 4 e)

Stecke

x = \frac{4 + 3 2}{2}

\frac{x ^{2} - 4}{e ^{x^{4} - 1}} : \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Maximum \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

Maximum - \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}, Minimum (0, - 4 e), Maximum \frac{4 + 3 2}{2}, \frac{3 2 - 4}{2 e ^{\frac{15 + 12 2}{2}}}

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

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

e x t re m e - 2 x^{3} + 14 x^{2} + 2 y^{2} + 4 x y

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

e x t re m e x + y