What is the extreme f(x)=3x^2+y^2-6x-4y+7 ?

The extreme f(x)=3x^2+y^2-6x-4y+7 is Minimum(1,2)

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extreme f(x)=3x^2+y^2-6x-4y+7

extreme points

f (x) = 3 x^{2} + y^{2} - 6 x - 4 y + 7

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

f (x) = 9 x^{10} - 3 x^{9}

28 x^{2} + 47 x + 15

- 2 x^{2} - 8 x + 2

\frac{x ^{5}}{20} - 4 x^{3}

y = x^{2} + 6 x + 5

What is the extreme f(x)=3x^2+y^2-6x-4y+7 ?
The extreme f(x)=3x^2+y^2-6x-4y+7 is Minimum(1,2)