The function given is:

$f(x, y) = 2x^3 - 6x + 6xy^2$

You are asked to find the local maximum, minimum values, and saddle points of the function. The image shows incorrect answers, so let's proceed with finding the correct points through critical point analysis and second partial derivative test.

Step 1: Find the partial derivatives of $f(x, y)$

Partial derivative with respect to $x$:

$f_x = \frac{\partial}{\partial x}(2x^3 - 6x + 6xy^2) = 6x^2 - 6 + 6y^2$

Partial derivative with respect to $y$:

$f_y = \frac{\partial}{\partial y}(2x^3 - 6x + 6xy^2) = 12xy$

Step 2: Solve for critical points by setting $f_x = 0$ and $f_y = 0$

• From $f_x = 6x^2 - 6 + 6y^2 = 0$, we have: $x^2 + y^2 = 1$

• From $f_y = 12xy = 0$, either $x = 0$ or $y = 0$.

Case 1: $x = 0$

Substituting $x = 0$ into $x^2 + y^2 = 1$, we get: $y^2 = 1 \implies y = \pm 1$ So, the critical points are $(0, 1)$ and $(0, -1)$.

Case 2: $y = 0$

Substituting $y = 0$ into $x^2 + y^2 = 1$, we get: $x^2 = 1 \implies x = \pm 1$ So, the critical points are $(1, 0)$ and $(-1, 0)$.

Thus, the critical points are $(0, 1)$, $(0, -1)$, $(1, 0)$, and $(-1, 0)$.

Step 3: Use the second partial derivative test

The second partial derivatives are: $f_{xx} = \frac{\partial}{\partial x}(6x^2 - 6 + 6y^2) = 12x$ $f_{yy} = \frac{\partial}{\partial y}(12xy) = 12x$ $f_{xy} = \frac{\partial}{\partial y}(6x^2 - 6 + 6y^2) = 12y$

The discriminant is: $D = f_{xx}f_{yy} - (f_{xy})^2$

Evaluating $D$ for each critical point:

1. At $(0, 1)$ and $(0, -1)$:

   • $f_{xx} = 0$, $f_{yy} = 0$, $f_{xy} = 12$.
   • $D = 0 \times 0 - (12)^2 = -144$ (negative, saddle points).

2. At $(1, 0)$ and $(-1, 0)$:

   • $f_{xx} = 12$, $f_{yy} = 12$, $f_{xy} = 0$.
   • $D = 12 \times 12 - 0 = 144$ (positive, check signs of $f_{xx}$).

Since $f_{xx} > 0$, both $(1, 0)$ and $(-1, 0)$ are local minima.

Final Answer:

• Local maximum: DNE
• Local minimum: $(1, 0)$, $(-1, 0)$
• Saddle points: $(0, 1)$, $(0, -1)$

Let me know if you need more details or have any questions!

5 Related Questions:

1. How do you classify critical points using the second partial derivative test?
2. What are the necessary conditions to determine a saddle point?
3. How can you visualize the behavior of the function around saddle points?
4. Why does $D < 0$ indicate a saddle point?
5. How can the function be graphed to highlight these critical points?

Tip:

Always check the discriminant $D$ carefully when classifying critical points—positive, negative, or zero gives important clues about the type of critical point.