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) = ln (27 + 8 x^{3})

f (x) = x^{5} + 10 x^{4} + 25 x^{3}

f (x, y) = x \cdot y

f (x) = 3 - x - \frac{2}{x}

f (x) = I x + 2 I + 3

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)