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 points

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

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})

f (2) = \frac{x}{e ^{- x}}

f (x) = 5 sin (x) cos (x), (- π, π)

y = j + x

f (x, y) = 2 x^{2} + y^{2} - 2 x y + 5 x - 3 y + 1

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

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)