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

Lösung

extrempunkte

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

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

Schritte zur Lösung

f^{'} (x) = e^{x^{3} - x} (3 x^{2} - 1)

Intervalle finden:

Ansteigend

: - \infty < x < - \frac{1}{3},

Absteigend

: - \frac{1}{3} < x < \frac{1}{3},

Ansteigend

: \frac{1}{3} < x < \infty

Stecke

x = - \frac{1}{3}

e^{x^{3} - x} : e^{\frac{2}{3 3}}

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

Stecke

x = \frac{1}{3}

e^{x^{3} - x} : \frac{1}{e ^{\frac{2}{3 3}}}

Minimum (\frac{1}{3}, \frac{1}{e ^{\frac{2}{3 3}}})

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

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

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

e x t re m e x^{5} - 5 x^{4} + 8

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

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