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

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

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

extreme points

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

Minimum (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Minimum (- 1, 0)

f (x) = 19 + 4 x - x^{2}

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

f (x) = e^{x} + e^{- x}

5 x + 9 x^{- 1}

f (x) = 3 x^{3} - 3 x^{2} + 12 x - 5, - 1 \leq x \leq 1

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