What is the extreme f(x)=x^3+y^2-3x-4y+7 ?

The extreme f(x)=x^3+y^2-3x-4y+7 is Saddle(-1,2),Minimum(1,2)

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extreme f(x)=x^3+y^2-3x-4y+7

extreme

Saddle (- 1, 2), Minimum (1, 2)

Solution steps

Find the critical points:

(- 1, 2), (1, 2)

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(1, 2) :

Minimum

Saddle (- 1, 2), Minimum (1, 2)

What is the extreme f(x)=x^3+y^2-3x-4y+7 ?
The extreme f(x)=x^3+y^2-3x-4y+7 is Saddle(-1,2),Minimum(1,2)