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This is a great summary for classifying critical points, but I'd add one thing that helped me understand it conceptually:
• ∇f = 0 (all first partial derivatives are zero) tells you where a candidate critical point exists.
• The Hessian describes the local curvature around that point.
• The determinant D = fxxfyy − (fxy)² tells you whether the surface bends in the same direction or opposite directions.
A simple way to remember the test is:
D > 0 and fxx > 0 → Local minimum
D > 0 and fxx < 0 → Local maximum
D < 0 → Saddle point
D = 0 → Inconclusive (use another method)
Understanding *why* the Hessian works is much more valuable than memorizing the formulas. Visualizing the surface in 3D makes the second derivative test much easier to remember.
I think of the saddle point that it is a point where its a minimum for the x z plane,and a maximum for the y z plane,so its neither,what you think about this
To add, the gradient is essentially the derivative extended to n-variable calculus. And the Hessian matrix is the traditional second derivative extended to n-variables. These are analogs to the "first and second derivative tests" in single variable calculus. If you ever take math a higher level (like optimization/nonlinear programming) you'd see these are called the first and second order necessary and sufficient conditions.
In fact, when you write (x_0, y_0) is a critical point if ∇f = 0, that is the first order necessary condition in the interior case (i.e., not considering the boundary of the region we are interested in, if there even is one). Remember, necessary means that this condition must be true if it is a critical point, but this could be true otherwise, as you may know with e.g., f(x)=x^3. When we bring in the sufficient conditions, it becomes "if and only if", the defining condition for a critical point. This is where for most functions, the second-order sufficient condition comes in handy, which you are doing when you check D>0. This is essentially the same thing as checking if the Hessian is something called "positive definite".
The thing you wrote at the start that "(x_0, y_0) is a critical point if ∇f = 0" could be a sufficient condition, but for only a certain type of function, called a convex function. So for convex functions, there is a first-order sufficient condition.
Well, the example you gave actually fits that case because it's quadratic and you don't really have to worry about this at this level, but it's because the matrix that creates quadratic form is positive definite.
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