What is the extreme f(x,y)=e^y-ye^x ?

The extreme f(x,y)=e^y-ye^x is Saddle(0,0)

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extreme f(x,y)=e^y-ye^x

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{y}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = \frac{x ^{3}}{( x + 1 )}

(x^{4} - 4 x^{3}), - 1 \leq x \leq 4

f (x) = x - 2 cos (x)

f (x) = \frac{x ^{2}}{x ^{2} + 4}

f (x) = x^{\frac{1}{3}} + 2 x^{\frac{4}{3}}

What is the extreme f(x,y)=e^y-ye^x ?
The extreme f(x,y)=e^y-ye^x is Saddle(0,0)