extreme y=(e^{(2-x)}+x^2-x)/2

Lösung

extrempunkte

y = \frac{e ^{(2 - x)} + x ^{2} - x}{2}

Minimum W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2}, \frac{4 W _{0} ( \frac{e ^{\frac{3}{2}}}{2} ) ^{2} + 4 e ^{\frac{3}{2} - W_{0} (\frac{e ^{\frac{3}{2}}}{2})} - 1}{8}

Schritte zur Lösung

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

Intervalle finden:

Absteigend

: - \infty < x < W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2},

Ansteigend

: W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2} < x < \infty

Stecke

x = W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2}

\frac{e ^{2 - x} + x ^{2} - x}{2} : \frac{4 W _{0} ( \frac{e ^{\frac{3}{2}}}{2} ) ^{2} + 4 e ^{\frac{3}{2} - W_{0} (\frac{e ^{\frac{3}{2}}}{2})} - 1}{8}

Minimum W_{0} (\frac{e ^{\frac{3}{2}}}{2}) + \frac{1}{2}, \frac{4 W _{0} ( \frac{e ^{\frac{3}{2}}}{2} ) ^{2} + 4 e ^{\frac{3}{2} - W_{0} (\frac{e ^{\frac{3}{2}}}{2})} - 1}{8}

e x t re m e f (x) = 20 (1 + \frac{1}{x} + \frac{1}{x ^{2}})

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

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

e x t re m e f (x, y) = 2 x^{2} + 2 y^{2} - 16 x + 12 y

e x t re m e f (x) = x^{3} - 15 x^{2} + 63 x - 15