# Find stationary points of a two variable function involving

## Homework Statement

Find all stationary points of the function

G(x, y) = (x^3)*e^(−x^2−y^2)

fx=0 and fy=0

## The Attempt at a Solution

Gx = 3x^2*e^(-x^2-y^2) +x^3(-2x)e^(-x^2-y^2) = e^(-x^2-y^2)(3x^2-2x^4)

Gx = 0 implies 3x^2-2x^4=0

x^2(3-2x^2)=0

hence x =0 ,+or- (3/2)^(1/2)

Gy = (-2y)(x^3)(e^(-x^2-y^2))

Gy=0

implies (-2y)(x^3)(e^(-x^2-y^2))=0

y=0 x = 0

(+or- (3/2)^(1/2), 0) are the two stationary points according to wolfram alpha but I don't understand why (o,a), where a is a real number, isn't a solution since Gx=Gy=0 when x = 0.

Any help would be appreciated.

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Ray Vickson
Science Advisor
Homework Helper
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## Homework Statement

Find all stationary points of the function

G(x, y) = (x^3)*e^(−x^2−y^2)

fx=0 and fy=0

## The Attempt at a Solution

Gx = 3x^2*e^(-x^2-y^2) +x^3(-2x)e^(-x^2-y^2) = e^(-x^2-y^2)(3x^2-2x^4)

Gx = 0 implies 3x^2-2x^4=0

x^2(3-2x^2)=0

hence x =0 ,+or- (3/2)^(1/2)

Gy = (-2y)(x^3)(e^(-x^2-y^2))

Gy=0

implies (-2y)(x^3)(e^(-x^2-y^2))=0

y=0 x = 0

(+or- (3/2)^(1/2), 0) are the two stationary points according to wolfram alpha but I don't understand why (o,a), where a is a real number, isn't a solution since Gx=Gy=0 when x = 0.

Any help would be appreciated.

The points (0,a) are solutions, but (according to you---I have not checked) Wolfram Alpha seems to have missed them. Maple did not miss them.

• Woolyabyss