What is the extreme f(x,y)=4x^2+y^2-8x+6y+8 ?

The extreme f(x,y)=4x^2+y^2-8x+6y+8 is Minimum(1,-3)

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extreme f(x,y)=4x^2+y^2-8x+6y+8

extreme points

f (x, y) = 4 x^{2} + y^{2} - 8 x + 6 y + 8

Minimum (1, - 3)

Solution steps

Find the critical points:

(1, - 3)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(1, - 3) :

Minimum

Minimum (1, - 3)

f (x) = e^{x^{2} - 9 x - 1}, - 9 \leq x \leq 9

f (x) = 5 - 7 x - 4 x^{2}

f (x) = ln (8 + x^{3})

f (x) = - x^{3} - 1

f (x, y) = 4 x y - x^{3} - 2 y^{2}

What is the extreme f(x,y)=4x^2+y^2-8x+6y+8 ?
The extreme f(x,y)=4x^2+y^2-8x+6y+8 is Minimum(1,-3)