# Find stationary points of a two variable function involving

• Woolyabyss
In summary, the function G(x, y) = (x^3)*e^(−x^2−y^2) has two stationary points at (+or- (3/2)^(1/2), 0), as shown by Wolfram Alpha. However, (0,a) where a is a real number, is also a solution as Gx=Gy=0 when x = 0. This solution may have been missed by Wolfram Alpha, but was identified by Maple.
Woolyabyss

## 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.

Last edited by a moderator:
Woolyabyss said:

## 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

## 1. What are stationary points?

Stationary points, also known as critical points, are points on a graph where the gradient or slope is equal to zero. This means that the tangent line at that point is horizontal, and the function is neither increasing nor decreasing.

## 2. How do I find stationary points of a two variable function?

To find the stationary points of a two variable function, you will need to take the partial derivatives of the function with respect to each variable and set them equal to zero. Then, solve the resulting system of equations to find the values of the variables at the stationary points.

## 3. Can a function have more than one stationary point?

Yes, a function can have multiple stationary points. These can be local maximums, local minimums, or saddle points. The number and type of stationary points will depend on the shape of the function's graph.

## 4. How do stationary points relate to the overall shape of a function's graph?

The stationary points of a function can give us information about the overall shape of its graph. For example, a local maximum or minimum stationary point can indicate a peak or valley in the graph, while a saddle point can indicate a change in concavity.

## 5. Why are stationary points important in mathematics and science?

Stationary points are important in mathematics and science because they help us understand the behavior of a function. They can give us information about the extrema, concavity, and other critical points of a function, which can be useful in optimization problems and other applications.

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