Intrinsic critical points of a function .

[Saint Medici]

Intrinsic critical points of a function...

I have a problem that's giving me a bit of trouble. The solution is not vital to my existence, but it's been eluding me long enough that it's grown to be a pest. I've figured out the easy part by hand, and I have the less obvious solutions by way of TI-89, but I'd like to know a way to get there without computer assistance. Anyway:

Find all the intrinsic critical points of f(x)=x_1^3+2x_1x_2x_3-x_3^2 on the unit sphere.

Of the ten, I've found six, namely, all the unit vectors along the coordinate axes. We've been using the Lagrange formulation quite a bit, so I assume that's what we're to use here. Anyway, any insight would be appreciated.

 

[wais]

Intrinsic critical points of a function refer to points where the gradient of the function is equal to zero. In other words, these are the points where the function is neither increasing nor decreasing in any direction. In this case, the function f(x) is defined on the unit sphere, which means that all the points (x_1, x_2, x_3) that satisfy the equation x_1^2 + x_2^2 + x_3^2 = 1 are considered.

To find the intrinsic critical points of f(x), we need to first calculate the gradient of the function. The gradient of a multivariable function is a vector that points in the direction of the steepest increase of the function at a given point. In this case, the gradient of f(x) is given by:

∇f(x) = (3x_1^2 + 2x_2x_3, 2x_1x_3, 2x_1x_2 - 2x_3)

Next, we set the gradient equal to zero and solve for the variables x_1, x_2, and x_3. This will give us the coordinates of the intrinsic critical points of f(x). In this case, we have:

3x_1^2 + 2x_2x_3 = 0
2x_1x_3 = 0
2x_1x_2 - 2x_3 = 0

From the second equation, we can see that either x_1 or x_3 must be equal to zero. Plugging this into the first and third equations, we get two possible solutions:

1) x_1 = 0, x_2 = 0, x_3 = ±1
2) x_1 = ±√(2/3), x_2 = 0, x_3 = ±√(1/3)

These solutions represent the intrinsic critical points of f(x) on the unit sphere. These points correspond to the unit vectors along the coordinate axes, as you have already found. So, your six solutions are correct.

In summary, to find the intrinsic critical points of a function, we need to calculate the gradient and set it equal to zero. Solving for the variables will give us the coordinates of the critical points. In this case, we found two sets of solutions