To analyze the function $f(x, y) = 3x^3 - 9x + 9xy^2$ and find its critical points, local maximum and minimum values, and any saddle points, we proceed as follows:

Step 1: Find the Partial Derivatives

To locate critical points, we need to calculate the partial derivatives $f_x$ and $f_y$ and set them to zero.

1. Partial derivative with respect to $x$: $f_x = \frac{\partial}{\partial x}(3x^3 - 9x + 9xy^2) = 9x^2 - 9 + 9y^2$

2. Partial derivative with respect to $y$: $f_y = \frac{\partial}{\partial y}(3x^3 - 9x + 9xy^2) = 18xy$

Step 2: Set Partial Derivatives to Zero

We solve the equations $f_x = 0$ and $f_y = 0$ simultaneously to find the critical points.

1. From $f_y = 18xy = 0$:

   • This gives $x = 0$ or $y = 0$.

2. Case 1: $x = 0$

   • Substitute $x = 0$ into $f_x = 9x^2 - 9 + 9y^2 = 0$: $-9 + 9y^2 = 0 \Rightarrow y^2 = 1 \Rightarrow y = \pm 1$
   • So, we have two critical points: $(0, 1)$ and $(0, -1)$.

3. Case 2: $y = 0$

   • Substitute $y = 0$ into $f_x = 9x^2 - 9 = 0$: $9x^2 - 9 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$
   • This gives two more critical points: $(1, 0)$ and $(-1, 0)$.

Step 3: Classify the Critical Points Using the Second Derivative Test

We need the second partial derivatives:

1. Second partial derivatives: $f_{xx} = \frac{\partial}{\partial x}(9x^2 - 9 + 9y^2) = 18x$ $f_{yy} = \frac{\partial}{\partial y}(18xy) = 18x$ $f_{xy} = \frac{\partial}{\partial y}(9x^2 - 9 + 9y^2) = 18y$

2. Calculate the Hessian Determinant, $D = f_{xx} f_{yy} - (f_{xy})^2$:

   • At $(0, 1)$: $f_{xx} = 0, \quad f_{yy} = 0, \quad f_{xy} = 18$ $D = (0)(0) - (18)^2 = -324$ Since $D < 0$, $(0, 1)$ is a saddle point.

   • At $(0, -1)$: $f_{xx} = 0, \quad f_{yy} = 0, \quad f_{xy} = -18$ $D = (0)(0) - (-18)^2 = -324$ Since $D < 0$, $(0, -1)$ is also a saddle point.

   • At $(1, 0)$: $f_{xx} = 18, \quad f_{yy} = 18, \quad f_{xy} = 0$ $D = (18)(18) - (0)^2 = 324$ Since $D > 0$ and $f_{xx} > 0$, $(1, 0)$ is a local minimum.

   • At $(-1, 0)$: $f_{xx} = -18, \quad f_{yy} = -18, \quad f_{xy} = 0$ $D = (-18)(-18) - (0)^2 = 324$ Since $D > 0$ and $f_{xx} < 0$, $(-1, 0)$ is a local maximum.

Step 4: Summarize the Results

• Local maximum value(s): $f(-1, 0) = 6$
• Local minimum value(s): $f(1, 0) = -6$
• Saddle point(s): $(0, 1, 0)$ and $(0, -1, 0)$

Final Answer

• Local maximum value(s): $6$
• Local minimum value(s): $-6$
• Saddle point(s): $(0, 1, 0), (0, -1, 0)$

Would you like details on graphing this function or further clarification?

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1. How do we interpret the Hessian determinant in classifying critical points?
2. What role do partial derivatives play in finding critical points?
3. Why are saddle points neither maximum nor minimum?
4. How does changing the function affect the number and type of critical points?
5. What are other methods to verify the nature of critical points besides the second derivative test?

Tip: Always verify critical points before applying the second derivative test to confirm calculation accuracy.