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

The extreme f(x,y)=3x^2+6xy+7y^2-2x+4y is Minimum(13/12 ,-3/4)

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

extreme points

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

Minimum (\frac{13}{12}, - \frac{3}{4})

Solution steps

Find the critical points:

(\frac{13}{12}, - \frac{3}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = 14

D (x, y) = 48

Check the sign of at each critical point

Check critical point

(\frac{13}{12}, - \frac{3}{4}) :

Minimum

Minimum (\frac{13}{12}, - \frac{3}{4})

f (x) = x^{\frac{1}{7}} - x^{- \frac{6}{7}}

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

f (x) = x^{\frac{1}{3}} (x^{2} - 9), - 4 \leq x \leq 2

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

f (x) = x^{4} - 4 x^{2} + 10

What is the extreme f(x,y)=3x^2+6xy+7y^2-2x+4y ?
The extreme f(x,y)=3x^2+6xy+7y^2-2x+4y is Minimum(13/12 ,-3/4)