What is the extreme f(x,y)=2e^{-y}(x^2+y^2)+5 ?

The extreme f(x,y)=2e^{-y}(x^2+y^2)+5 is Minimum(0,0),Saddle(0,2)

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

extreme points

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

Minimum (0, 0), Saddle (0, 2)

Solution steps

Find the critical points:

(0, 0), (0, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 (e^{- y} y^{2} - 4 e^{- y} y + e^{- y} x^{2} + 2 e^{- y})

D (x, y) = - 8 e^{- 2 y} x^{2} + 16 e^{- 2 y} + 8 e^{- 2 y} y^{2} - 32 e^{- 2 y} y

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(0, 2) :

Saddle

Minimum (0, 0), Saddle (0, 2)

f (x) = \frac{5 - x ^{2}}{x}

x^{2} - 14 x + 2

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

y = 2 x^{2} - 5 x + 4 - 10

x^{4} - 4 x^{3} - 18 x^{2} - 20 x - 12

What is the extreme f(x,y)=2e^{-y}(x^2+y^2)+5 ?
The extreme f(x,y)=2e^{-y}(x^2+y^2)+5 is Minimum(0,0),Saddle(0,2)