What is the extreme f(x,y)=x^{e-2x^2-2y^2} ?

The extreme f(x,y)=x^{e-2x^2-2y^2} is Saddle(1,sqrt((e-2)/2)),Saddle(1,-sqrt((e-2)/2))

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extreme f(x,y)=x^{e-2x^2-2y^2}

extreme points

f (x, y) = x^{e - 2 x^{2} - 2 y^{2}}

Saddle (1, \frac{e - 2}{2}), Saddle (1, - \frac{e - 2}{2})

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 ln (x) (x^{e - 2 x^{2} - 2 y^{2}} - 4 y^{2} x^{e - 2 x^{2} - 2 y^{2}} ln (x))

D (x, y) = - 4 x^{- 2 x^{2} - 2 y^{2} + e} ln (x) (- 4 (1 + ln (x)) + \frac{- e - 2 x ^{2} + 2 y ^{2} + ( e - 2 x ^{2} - 2 y ^{2} ) ^{2} - 8 e x ^{2} ln ( x ) + 16 x ^{4} ln ( x ) + 16 x ^{2} y ^{2} ln ( x ) + 16 x ^{4} ln ^{2} ( x )}{x ^{2}}) (x^{e - 2 x^{2} - 2 y^{2}} - 4 y^{2} x^{e - 2 x^{2} - 2 y^{2}} ln (x)) - 16 y^{2} x^{2 (- 2 x^{2} - 2 y^{2} + e)} (\frac{( ln ( x ) ( e - 2 x ^{2} - 2 y ^{2} ) + 1 ) ^{2}}{x ^{2}} - 8 e ln^{3} (x) + 16 x^{2} ln^{3} (x) + 16 y^{2} ln^{3} (x) - 8 ln^{2} (x) + 16 x^{2} ln^{4} (x))

Check the sign of at each critical point

Check critical point

(1, \frac{e - 2}{2}) :

Saddle

Check critical point

(1, - \frac{e - 2}{2}) :

Saddle

Saddle (1, \frac{e - 2}{2}), Saddle (1, - \frac{e - 2}{2})

f (x) = x ln (\frac{x}{3}), 0.1 \leq x \leq 3

f (x) = 2 x^{3} - 6 x^{2} + 6 x - 8

x^{3} - 27 x + 51

x^{3} - 27 x + 57

f (x) = 2 x^{3} - 6 x^{2} + 6 x + 1

What is the extreme f(x,y)=x^{e-2x^2-2y^2} ?
The extreme f(x,y)=x^{e-2x^2-2y^2} is Saddle(1,sqrt((e-2)/2)),Saddle(1,-sqrt((e-2)/2))