https://file.aiquickdraw.com/mathbot/file/172971188399334lrzj8r.jpg

The image presents a problem asking to find the local maximum, minimum, and saddle points for a function \( f(x, y) = x^4 + y^4 - 4xy + 1 \).

### Steps to solve the problem:

1. **First Derivatives**: To locate critical points, we need to find the first partial derivatives of \( f(x, y) \) and set them equal to zero.

   \[
   f_x = \frac{\partial}{\partial x} (x^4 + y^4 - 4xy + 1) = 4x^3 - 4y
   \]
   \[
   f_y = \frac{\partial}{\partial y} (x^4 + y^4 - 4xy + 1) = 4y^3 - 4x
   \]

   Now, solve the system of equations:
   \[
   4x^3 - 4y = 0 \quad \text{(1)}
   \]
   \[
   4y^3 - 4x = 0 \quad \text{(2)}
   \]

2. **Solve the System**:
   From equation (1), simplify to:
   \[
   x^3 = y
   \]
   From equation (2), simplify to:
   \[
   y^3 = x
   \]
   Substitute \( y = x^3 \) into \( x = y^3 \) to solve for \( x \):
   \[
   x = (x^3)^3 = x^9
   \]
   This gives two possible solutions:
   \[
   x = 0 \quad \text{or} \quad x = 1 \quad \text{or} \quad x = -1
   \]

   - For \( x = 0 \), substitute into \( y = x^3 \): \( y = 0 \).
   - For \( x = 1 \), substitute into \( y = x^3 \): \( y = 1 \).
   - For \( x = -1 \), substitute into \( y = x^3 \): \( y = -1 \).

   So, the critical points are \( (0, 0) \), \( (1, 1) \), and \( (-1, -1) \).

3. **Second Derivatives and Hessian**:
   To classify the critical points, we compute the second partial derivatives:
   \[
   f_{xx} = 12x^2, \quad f_{yy} = 12y^2, \quad f_{xy} = -4
   \]

   The Hessian matrix \( H \) is:
   \[
   H = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{bmatrix} = \begin{bmatrix} 12x^2 & -4 \\ -4 & 12y^2 \end{bmatrix}
   \]
   The determinant of the Hessian is:
   \[
   \det(H) = (12x^2)(12y^2) - (-4)(-4) = 144x^2y^2 - 16
   \]

4. **Classify Critical Points**:
   - At \( (0, 0) \), \( \det(H) = 144(0)(0) - 16 = -16 \), so \( (0, 0) \) is a saddle point.
   - At \( (1, 1) \), \( \det(H) = 144(1)(1) - 16 = 128 \), and since \( f_{xx} = 12 > 0 \), \( (1, 1) \) is a local minimum.
   - At \( (-1, -1) \), \( \det(H) = 144(1)(1) - 16 = 128 \), and since \( f_{xx} = 12 > 0 \), \( (-1, -1) \) is a local minimum.

### Conclusion:
- **Saddle point**: \( (0, 0) \)
- **Local minima**: \( (1, 1) \), \( (-1, -1) \)

Let me know if you'd like more details or have any questions! Here are some further exploration points:

1. How do we interpret saddle points geometrically?
2. What would change if the function had higher-degree terms?
3. Can you find the global minimum for this function?
4. What happens if we add constraints like \( x + y = 0 \)?
5. How can we visualize the surface of this function?

**Tip**: The Hessian matrix is a powerful tool for classifying critical points in multivariable functions.

Find the local maximum, minimum, and saddle points for the function f(x, y) = x^4 + y^4 - 4xy + 1.

Learn how to find the local maxima, minima, and saddle points for the function f(x, y) = x^4 + y^4 - 4xy + 1. This problem involves calculating first and second derivatives, solving critical points, and using the Hessian matrix for classification. Suitable for undergraduate calculus students.

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