Given that $x^*$ is a stationary point and the Hessian matrix at $x^*$ is positive semidefinite, we can analyze the nature of the stationary point using the properties of the Hessian matrix.

Key points:

1. Stationary point: At a stationary point $x^*$, the gradient of the function $f$ is zero, i.e., $\nabla f(x^*) = 0$.
2. Positive semidefinite Hessian: If the Hessian matrix $H(x^*)$ is positive semidefinite, then for any vector $v$, we have: $v^T H(x^*) v \geq 0$ This means that the second-order directional derivatives are non-negative in every direction.

Interpretation:

• A positive semidefinite Hessian suggests that the curvature is either positive or zero in all directions at $x^*$.
• However, positive semidefinite does not imply strict positivity in all directions (which would be positive definite). It allows for the possibility that the Hessian has zero eigenvalues in some directions.
• Therefore, $x^*$ could be a local minimizer if the Hessian is strictly positive in certain directions, but if there are directions where the Hessian is zero, it could also be a saddle point.

Thus, $x^*$ can be:

• A local minimizer.
• A saddle point.
• It cannot be a local maximizer, because a positive semidefinite Hessian does not permit the necessary concavity for a local maximum.

Conclusion:

The correct statement is:

• $x^*$ can be a local minimizer, local maximizer, or saddle point.

This option corresponds to the following choice:

• x* can be a local minimizer, local maximizer, or saddle point.

Would you like further details on how the Hessian helps classify stationary points?

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5 Relative Questions:

1. How does the definiteness of the Hessian affect the classification of stationary points?
2. What role does the second derivative test play in identifying local extrema?
3. Can a positive semidefinite Hessian guarantee convexity of the function?
4. How does a positive definite Hessian differ from a positive semidefinite Hessian in terms of classification of points?
5. What are the implications if the Hessian is negative semidefinite at a stationary point?

Tip: The eigenvalues of the Hessian matrix can provide important insights into the local curvature of the function and the nature of stationary points.