What is the extreme f(x)=2x^2-4x+y^2-6y+3 ?

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

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extreme f(x)=2x^2-4x+y^2-6y+3

extreme points

f (x) = 2 x^{2} - 4 x + y^{2} - 6 y + 3

Minimum (1, 3)

Solution steps

Find the critical points:

(1, 3)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, 3) :

Minimum

Minimum (1, 3)

\frac{x + 2}{x ^{2} - x - 6}

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

f (x) = 2 x^{3} + 9 x^{2} + 12 x - 9

f (x, y) = 7 x^{2} - x y + 10 y^{2}

f (x) = x^{3} - 3 x^{2} - 105 x

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