What is the extreme f(x,y)=e^{-x-y^2}*(3x-y^2) ?

The extreme f(x,y)=e^{-x-y^2}*(3x-y^2) is Maximum(1,0)

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extreme f(x,y)=e^{-x-y^2}*(3x-y^2)

extreme points

f (x, y) = e^{- x - y^{2}} \cdot (3 x - y^{2})

Maximum (1, 0)

Solution steps

Find the critical points:

(1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 (2 e^{- y^{2} - x} y^{4} - 6 e^{- y^{2} - x} x y^{2} - 5 e^{- y^{2} - x} y^{2} + 3 e^{- y^{2} - x} x + e^{- y^{2} - x})

D (x, y) = 2 e^{- y^{2} - x} (y^{2} + 6 - 3 x) (2 e^{- y^{2} - x} y^{4} - 6 e^{- y^{2} - x} x y^{2} - 5 e^{- y^{2} - x} y^{2} + 3 e^{- y^{2} - x} x + e^{- y^{2} - x}) - 4 e^{2 (- y^{2} - x)} y^{2} (3 x - y^{2} - 2)^{2}

Check the sign of at each critical point

Check critical point

(1, 0) :

Maximum

Maximum (1, 0)

x^{2} + y^{2} + 2 x - 4 y + 8

f (x) = \frac{( 3 x - 6 )}{( x + 2 )}

f (x, y) = 3 x^{3} - y^{3} - 9 x + 12 y + 3

f (x) = \frac{x}{5} + \frac{4 + ( 6 - x ) ^{2}}{3}

f (x) = - x^{3} + 3 x^{2} + 9 x - 1

What is the extreme f(x,y)=e^{-x-y^2}*(3x-y^2) ?
The extreme f(x,y)=e^{-x-y^2}*(3x-y^2) is Maximum(1,0)