What is the extreme f(x,y)=2x^2+3y^2-4x-8 ?

The extreme f(x,y)=2x^2+3y^2-4x-8 is Minimum(1,0)

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extreme f(x,y)=2x^2+3y^2-4x-8

extreme points

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 8

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

13 + 4 x - x^{2}

f (x) = \frac{ln ( x )}{x ^{3}}

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

f (x) = x^{4} - 18 x^{2} + 11

f (x, y) = - 5 x^{2} y + 5 x y^{2}

What is the extreme f(x,y)=2x^2+3y^2-4x-8 ?
The extreme f(x,y)=2x^2+3y^2-4x-8 is Minimum(1,0)