The function

f(x,y)=x3−3x−2xy−y2−2y+4

has two critical points. Find and classify them. 

Note: For non-integer numerical values, you must use at least 3 decimal places. Also, you must use a 'full stop' . and not a comma ',' for a decimal point.

The left-most stationary point (the one with the lower value of x
) is

located at x=
Answer 1 Question 3
,   y=
Answer 2 Question 3
, and is a Answer 3 Question 3
 

The right-most stationary point (the one with the higher value of x
) is

located at x=
Answer 4 Question 3
,   y=
Answer 5 Question 3
, and is a Answer 6 Question 3

To find the critical points of the function \( f(x,y) = x^3 - 3x - 2xy - y^2 - 2y + 4 \), we first need to compute the partial derivatives with respect to \( x \) and \( y \), and then set them to zero.

### Step 1: Compute the partial derivatives
The partial derivative of \( f(x, y) \) with respect to \( x \) is:
\[
\frac{\partial f}{\partial x} = 3x^2 - 3 - 2y
\]

The partial derivative of \( f(x, y) \) with respect to \( y \) is:
\[
\frac{\partial f}{\partial y} = -2x - 2y - 2
\]

### Step 2: Set the partial derivatives equal to zero
To find the critical points, solve the system of equations:
\[
3x^2 - 3 - 2y = 0
\]
\[
-2x - 2y - 2 = 0
\]

### Step 3: Solve the system of equations

1. From the second equation:
   \[
   -2x - 2y - 2 = 0
   \]
   Simplifying this:
   \[
   x + y + 1 = 0 \quad \text{or} \quad y = -x - 1
   \]

2. Substitute \( y = -x - 1 \) into the first equation:
   \[
   3x^2 - 3 - 2(-x - 1) = 0
   \]
   Simplifying:
   \[
   3x^2 - 3 + 2x + 2 = 0 \quad \text{or} \quad 3x^2 + 2x - 1 = 0
   \]

3. Solve the quadratic equation \( 3x^2 + 2x - 1 = 0 \) using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 3 \), \( b = 2 \), and \( c = -1 \):
   \[
   x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-1)}}{2(3)} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6}
   \]
   This gives two solutions for \( x \):
   \[
   x_1 = \frac{2}{6} = \frac{1}{3} \approx 0.333
   \]
   \[
   x_2 = \frac{-6}{6} = -1
   \]

4. Find the corresponding \( y \) values:
   - For \( x_1 = 0.333 \):
     \[
     y_1 = -0.333 - 1 \approx -1.333
     \]
   - For \( x_2 = -1 \):
     \[
     y_2 = -(-1) - 1 = 0
     \]

### Step 4: Classify the critical points

To classify the critical points, compute the second partial derivatives and evaluate the Hessian determinant.

1. Compute the second partial derivatives:
   \[
   f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x
   \]
   \[
   f_{yy} = \frac{\partial^2 f}{\partial y^2} = -2
   \]
   \[
   f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -2
   \]

2. Compute the Hessian determinant \( D \) at each critical point:
   \[
   D = f_{xx}f_{yy} - (f_{xy})^2
   \]

- **At \( (x_1, y_1) = (0.333, -1.333) \):**
  \[
  D_1 = 6(0.333) \times (-2) - (-2)^2 = -4 - 4 = -8
  \]
  Since \( D_1 < 0 \), \( (0.333, -1.333) \) is a **saddle point**.

- **At \( (x_2, y_2) = (-1, 0) \):**
  \[
  D_2 = 6(-1) \times (-2) - (-2)^2 = 12 - 4 = 8
  \]
  Since \( D_2 > 0 \) and \( f_{xx} < 0 \), \( (-1, 0) \) is a **local maximum**.

### Final Answers:
- The left-most stationary point is located at \( x = -1 \), \( y = 0 \), and is a **local maximum**.
- The right-most stationary point is located at \( x = 0.333 \), \( y = -1.333 \), and is a **saddle point**.

Would you like more details or have any questions?

---

**Questions to explore:**
1. What is the significance of a saddle point in optimization problems?
2. How do you interpret the Hessian determinant in the context of multivariable calculus?
3. What are the potential applications of finding critical points in real-world problems?
4. How do local maxima and minima relate to the global maximum and minimum of a function?
5. What happens if the Hessian determinant is zero at a critical point?
6. Can a critical point be neither a maximum, minimum, nor a saddle point?
7. How would you solve this problem graphically?
8. What role does the quadratic formula play in solving higher-order equations?

**Tip:** When analyzing critical points, always check the signs of both the Hessian determinant and the second partial derivatives to classify the points correctly.

Learn how to find and classify critical points of the function f(x, y) = x^3 - 3x - 2xy - y^2 - 2y + 4 using multivariable calculus methods. Explore detailed solutions, including the application of the Second Derivative Test to determine the nature of critical points.

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