What is the extreme f(x,y)=x^2-2y^2+8x-3y+1 ?

The extreme f(x,y)=x^2-2y^2+8x-3y+1 is Saddle(-4,-3/4)

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extreme f(x,y)=x^2-2y^2+8x-3y+1

extreme

Saddle (- 4, - \frac{3}{4})

Solution steps

Find the critical points:

(- 4, - \frac{3}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = - 8

Check the sign of at each critical point

Check critical point

(- 4, - \frac{3}{4}) :

Saddle

Saddle (- 4, - \frac{3}{4})

What is the extreme f(x,y)=x^2-2y^2+8x-3y+1 ?
The extreme f(x,y)=x^2-2y^2+8x-3y+1 is Saddle(-4,-3/4)