Question

Consider the function \( f(x,y) = (x - 2)^2(y + 3) \). Then:

• A. (2,-3) is not a stationary point of \( f \).
• B. \( f \) has a local minimum at (2,-3).
• C. \( f \) has a local maximum at (2,-3).
• D. \( f \) has neither a local maximum nor a local minimum at (2,-3).

Hint:

If the second derivative test \( D = f_{xx}f_{yy} - (f_{xy})^2 \) equals 0, the test is inconclusive. In such cases, look at the sign of the function relative to its value at the stationary point to identify saddle points.

Answer 1

The Correct Option is D

Concept:
A point \( (x_0, y_0) \) is a stationary point if the partial derivatives \( f_x \) and \( f_y \) are zero. To classify it, we look at the second derivative test or the behavior of the function in the neighborhood of the point.

Step 1: Find the partial derivatives.
\[ f_x = \frac{\partial}{\partial x} [ (x-2)^2(y+3) ] = 2(x-2)(y+3) \] \[ f_y = \frac{\partial}{\partial y} [ (x-2)^2(y+3) ] = (x-2)^2 \] At the point \( (2, -3) \): \[ f_x(2, -3) = 2(0)(0) = 0 \] \[ f_y(2, -3) = (0)^2 = 0 \] Thus, \( (2, -3) \) is a stationary point.

Step 2: Analyze the neighborhood of the stationary point.
Note that \( f(2, -3) = (0)^2(0) = 0 \). Let's check the value of \( f(x, y) \) very close to \( (2, -3) \): The term \( (x-2)^2 \) is always \( \geq 0 \) for any \( x \neq 2 \).
• If \( y > -3 \), then \( (y+3) > 0 \), so \( f(x, y) > 0 \).
• If \( y < -3 \), then \( (y+3) < 0 \), so \( f(x, y) < 0 \).

Step 3: Conclusion on extrema.
In any neighborhood around \( (2, -3) \), the function takes both positive and negative values. Since \( f(2, -3) = 0 \), there are points arbitrarily close that are both larger than 0 and smaller than 0. Therefore, the function has a saddle point behavior at \( (2, -3) \).