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

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

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

extreme points

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

Minimum (1, - 1)

Solution steps

Find the critical points:

(1, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(1, - 1) :

Minimum

Minimum (1, - 1)

\frac{- 8 x + 75}{9 x - 61}

4 + (7 + 5 x)^{\frac{2}{5}}

2.8 x

f (x) = - \frac{x ^{2} - 50}{x ^{2} - 25}

f (x) = (\frac{1}{8}) e^{- x}

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