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

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

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extreme f(x)=3x^2-2x+y^2-4y+1

extreme points

f (x) = 3 x^{2} - 2 x + y^{2} - 4 y + 1

Minimum (\frac{1}{3}, 2)

Solution steps

Find the critical points:

(\frac{1}{3}, 2)

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

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(\frac{1}{3}, 2) :

Minimum

Minimum (\frac{1}{3}, 2)

P (a, b) = 2 a + b

y = x^{2} - 2 ln (x)

f (x) = - e^{\frac{2}{7} x}

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

f (x) = \frac{4 x - 8}{x ^{3}}

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