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

The extreme f(x,y)=2x^2+3y^2-7 is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{x ^{2} + 4}{x}

f (x, y) = x^{3} - 6 x y + 3 y^{2} + 1

f (x) = \frac{x ^{2}}{x - 6}

f (x, y) = 4 x^{3} - 11 x y^{2} + 15 y + 12

f (x) = sin (x) + cos^{2} (x)

What is the extreme f(x,y)=2x^2+3y^2-7 ?
The extreme f(x,y)=2x^2+3y^2-7 is Minimum(0,0)