Question

Let f : \(\R^2 → \R\) be the function defined by
f(x, y) = 8(x² - y²) - x⁴ + y⁴.
Then which of the following statements is/are true ?

• A. f has 9 critical points
• B. f has a saddle point at (2, 2)
• C. f has a local maximum at (−2, 0)
• D. f has a local minimum at (0, −2)

Answer 1

The Correct Option is A, B, C, D

To determine the correct statements about the function \(f(x, y) = 8(x^2 - y^2) - x^4 + y^4\), we need to analyze the function for its critical points, nature of these critical points, and local extrema. Let's proceed with the analysis step-by-step:

Finding Critical Points:
Critical points occur where the first partial derivatives are zero.

Compute the partial derivatives:

\(\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left[ 8(x^2 - y^2) - x^4 + y^4 \right] = 16x - 4x^3\)

\(\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left[ 8(x^2 - y^2) - x^4 + y^4 \right] = -16y + 4y^3\)

Setting these partial derivatives to zero gives:

\(16x - 4x^3 = 0\) 
Factor out 4x: \(4x(4 - x^2) = 0\)

This gives \(x = 0\) or \(x = \pm 2\).

\(-16y + 4y^3 = 0\) 
Factor out 4y: \(4y(y^2 - 4) = 0\)

This gives \(y = 0\) or \(y = \pm 2\).

Thus, the critical points are: \((0, 0)\), \((0, 2)\), \((0, -2)\), \((2, 0)\), \((2, 2)\), \((2, -2)\), \((-2, 0)\), \((-2, 2)\), \((-2, -2)\). Therefore, there are 9 critical points.

Evaluating Nature of Critical Points: Saddle Point, Maximum, Minimum

Use the second derivative test. Compute the second partial derivatives:

\(\frac{\partial^2 f}{\partial x^2} = 16 - 12x^2\)

\(\frac{\partial^2 f}{\partial y^2} = -16 + 12y^2\)

\(\frac{\partial^2 f}{\partial x \partial y} = 0\)

Evaluate the determinant of the Hessian matrix \(H(x, y)\) at each point:

\(H(x, y) = \begin{pmatrix} \frac{\partial^2 f}{\partial x^2} & 0 \\ 0 & \frac{\partial^2 f}{\partial y^2} \end{pmatrix}\)

The determinant is \(\det(H) = \left(16 - 12x^2\right)\left(-16 + 12y^2\right) - 0 = (16 - 12x^2)(-16 + 12y^2)\).

• At (2, 2): \(\det(H) = (16 - 12(2)^2)(-16 + 12(2)^2) = (16 - 48)(32 - 16) = (-32)(16) < 0\), indicating a saddle point.
• At (−2, 0): \(\det(H) = (16 - 48)(-16 + 0) = (-32)(-16) > 0\), and \(\frac{\partial^2 f}{\partial x^2} < 0\), indicating a local maximum.
• At (0, −2): \(\det(H) = (16 - 0)(-16 + 48) = (16)(32) > 0\), and \(\frac{\partial^2 f}{\partial x^2} > 0\), indicating a local minimum.

Conclusion:

• f has 9 critical points.
• f has a saddle point at (2, 2).
• f has a local maximum at (−2, 0).
• f has a local minimum at (0, −2).

Thus, all the given options are true.