❓ Submitted Question: Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) f(x, y) = x4 + y4 − 4xy + 2

🧠Answer:

To find the critical points, we need to find where the gradient of the function equals the zero vector:

∇f(x, y) = ⟨4x3 - 4y, 4y3 - 4x⟩ = ⟨0, 0⟩

This gives us the system of equations:

4x3 - 4y = 0

4y3 - 4x = 0

Simplifying, we get:

y = x3

y3 = x

Substituting y = x3 into y3 = x, we get:

x9 = x

x(x8 - 1) = 0

The solutions to this equation are x = 0 and x = ±1, which gives us the critical points:

(0, 0) and (±1, ±1)

To determine the nature of these critical points, we need to calculate the second partial derivatives of f:

fxx = 12x2, fxy = fyx = -4, fyy = 12y2

At the critical point (0, 0):

fxx(0, 0) = 0, fxy(0, 0) = -4, fyy(0, 0) = 0

The discriminant of the Hessian matrix is:

D = fxx(0, 0)fyy(0, 0) - fxy(0, 0)2 = 0 - (-4)2 = 16

Since D > 0 and fxx(0, 0) = fyy(0, 0) = 0, the second derivative test is inconclusive. We need to use another method to determine the nature of this critical point.

One way to do this is to look at the behavior of f along certain curves. For example, if we look at the curve y = x4, then f(x, x4) = 2 for all x. This tells us that (0, 0) is a saddle point.

At the critical point (1, 1):

fxx(1, 1) = 12, fxy(1, 1) = fyx(1, 1) = -4, fyy(1, 1) = 12

The discriminant of the Hessian matrix is:

D = fxx(1, 1)fyy(1, 1) - fxy(1, 1)2 = 144 - (-4)2 = 128

Since D > 0 and fxx(1, 1) > 0, the second derivative test tells us that (1, 1) is a local minimum.

Similarly, at the critical point (-1, -1), we can use the second derivative test to show that it is a local minimum.

At the critical point (-1, 1), we can look at the curve y = -x4 + 2 to see that f(-1, 1) = f(1, -1) = 2. This tells us that (-1, 1) and (1, -1) are saddle points.

Therefore, the local maximum and minimum values are DNE, and the saddle points are (0, 0), (-1, 1), and (1, -1).

To graph the function, we can use a three-dimensional graphing software. Here is a graph of f(x, y) = x4 + y4 − 4xy + 2:

![Graph of f(x, y) = x^4 + y^4 − 4xy + 2](https://i.imgur.com/1zfuKbr.png

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