What is the extreme (x^2-1)*(e^y-1)+(y^2-2y+2)e^y ?

The extreme (x^2-1)*(e^y-1)+(y^2-2y+2)e^y is Minimum(0,1),Maximum(0,-1),Saddle(1,0),Saddle(-1,0)

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extreme (x^2-1)*(e^y-1)+(y^2-2y+2)e^y

extreme

Minimum (0, 1), Maximum (0, - 1), Saddle (1, 0), Saddle (- 1, 0)

Solution steps

Find the critical points:

(0, 1), (0, - 1), (1, 0), (- 1, 0)

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

D (x, y) = 2 e^{2 y} y^{2} - 2 e^{y} y^{2} + 4 e^{2 y} y - 4 e^{y} y - 2 e^{2 y} x^{2} - 2 e^{y} x^{2} - 2 e^{2 y} + 2 e^{y}

Check the sign of at each critical point

Check critical point

(0, 1) :

Minimum

Check critical point

(0, - 1) :

Maximum

Check critical point

(1, 0) :

Saddle

Check critical point

(- 1, 0) :

Saddle

Minimum (0, 1), Maximum (0, - 1), Saddle (1, 0), Saddle (- 1, 0)

What is the extreme (x^2-1)*(e^y-1)+(y^2-2y+2)e^y ?
The extreme (x^2-1)*(e^y-1)+(y^2-2y+2)e^y is Minimum(0,1),Maximum(0,-1),Saddle(1,0),Saddle(-1,0)