# Finding local extremes from stationary points

• goatsebear
In summary, the stationary points and local extreme values of f(x,y) = xy^2e^-(\frac{x^2 + y^2)}{2} are (0,0), (1,-\sqrt{2}) and (1, -\sqrt{2}).
goatsebear

## Homework Statement

Find the stationary points and local extreme values of f(x,y) = xy^2e^-($$\frac{x^2 + y^2)}{2}$$

## Homework Equations

You need to find the gradient for the function and then set it equal to 0. So then df/dy equals df/dx and you can solve for a solution set.

## The Attempt at a Solution

df/dx = y^2(-x^2 + 1)*e^-($$\frac{x^2 + y^2)}{2}$$

df/dy = y*x(-y^2 + 2)*e^-($$\frac{x^2 + y^2)}{2}$$

So since e^-($$\frac{x^2 + y^2)}{2}$$ cannot equal zero, y^2(-x^2 + 1) = 0
and y*x(-y^2 + 2) = 0

So then, y^2(-x^2 + 1) goes to y^2 - x^2y^2 = 0 and then $$\pm$$1 = x

Plugging this into the dy/dx equation gives me the following x, y critical points
(0,0) (1,-$$\sqrt{2}$$) (1, -$$\sqrt{2}$$) (-1, $$\sqrt{2}$$) (-1, $$\sqrt{-2}$$)

So then to find out if they are max or mins or saddle points, I solve for the second partials of fx, fy, fxy

fxx=A = x(-1 + 2y^2 - x^2y^2) * e^-($$\frac{x^2 + y^2)}{2}$$

fyy=C = y(x - 2y + 3yx) * e^-($$\frac{x^2 + y^2)}{2}$$

fxy=B = y(2- 2x^2 + y^2 = Y2x^2) * e^-($$\frac{x^2 + y^2)}{2}$$Now I use the equation D = AC - B^2

Can someone check my math out? It seems that something is wrong somewhere. I know this is a huge problem so thanks for any help.

Last edited:
The critical points are:
(x, 0) for any x
(-1, -sqrt 2)
(-1, sqrt 2)
(1, -sqrt 2)
(1, sqrt 2)

In each 2nd order derivative, the polynomial term should simplify to:
x(-3 + x^2)y^2 ... for "xx"
x(2 - 5 y^2 + y^4) ... for "yy"
y(-1 + x^2)(-2 + y^2) ... for "xy"

Last edited:
So I was mostly right other than that being (x,0). Now I just plug those points in for x and y in that D = AC - B^2 right? How would I do that with the (x,0) point?

Plug in y = 0, leave x as x.

## 1. How do stationary points help in finding local extremes?

Stationary points, also known as critical points, are points where the derivative of a function is equal to zero. Local extremes occur at these points because the derivative changes from positive to negative or vice versa. By finding the stationary points, we can determine where the function changes from increasing to decreasing or vice versa, indicating the presence of a local extreme.

## 2. What is the difference between a local extreme and a global extreme?

A local extreme is a point where a function reaches a maximum or minimum value within a specific interval. A global extreme, on the other hand, is the highest or lowest point of the entire function. In other words, a local extreme occurs within a certain range, while a global extreme encompasses the entire function.

## 3. How do you determine if a stationary point is a local maximum or minimum?

To determine if a stationary point is a local maximum or minimum, we can use the first or second derivative test. The first derivative test involves evaluating the sign of the derivative on either side of the stationary point. If the sign changes from positive to negative, the point is a local maximum. If the sign changes from negative to positive, the point is a local minimum. The second derivative test involves evaluating the concavity of the graph at the stationary point. If the second derivative is positive, the point is a local minimum. If the second derivative is negative, the point is a local maximum.

## 4. Can a stationary point be neither a local maximum nor minimum?

Yes, a stationary point can also be an inflection point, where the function changes concavity but does not have a local extreme. This occurs when the second derivative is equal to zero at the stationary point.

## 5. Are there any other methods for finding local extremes besides using stationary points?

Yes, there are other methods such as using the Mean Value Theorem, Rolle's Theorem, and the Intermediate Value Theorem. These theorems can provide information about the behavior of a function and help in identifying local extremes. Additionally, graphical methods such as using a graphing calculator or plotting the function by hand can also aid in finding local extremes.

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