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

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

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

extreme points

f (x) = 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{1}{3} x^{3} - x^{2} + x - 5

f (x) = 2 x^{2} + 3 x y + 4 y^{2} - 5 x + 2 y

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

f (x) = - x^{2} + 4 x + 9

f (x) = x^{- 2} ln (x)

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