What is the extreme f(x)=3x^2+2y^2-18x+4y ?

The extreme f(x)=3x^2+2y^2-18x+4y is Minimum(3,-1)

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extreme f(x)=3x^2+2y^2-18x+4y

extreme points

f (x) = 3 x^{2} + 2 y^{2} - 18 x + 4 y

Minimum (3, - 1)

Solution steps

Find the critical points:

(3, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(3, - 1) :

Minimum

Minimum (3, - 1)

\frac{2 x - 3}{x - 2}

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

f (x) = - 2 x^{3} + 9 x^{2} + 168 x + 69

U (x, y) = (x + \frac{x}{y}) (2 y + 3)

- e^{- x} x

What is the extreme f(x)=3x^2+2y^2-18x+4y ?
The extreme f(x)=3x^2+2y^2-18x+4y is Minimum(3,-1)