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I have a function of which I have identified a series of critical point.


The partial derivates

$A = \dfrac{d}{dxdx}f(x,y)=0 $

$B = \dfrac{d}{dxdy}f(x,y)=0 $

$C = \dfrac{d}{dydy}f(x,y)=2y+\cos(y)$

This gives me a series of critical points at $(x,-0.45)$

I don't know how to analytical answer if those are saddle points, minimum or maximum as the formula $B^2-A*C = 0 = $no information.

However using numerical computer programs I can see that it is a minimum point. However I don't know if this is an acceptable approached in univeristy? Graph

enter image description here

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Note that $f$ is independent of $x$, so any minimum will be global, but cannot be a local minimum--that is, there is no neighborhood of the point in which it is the only point attaining the minimum value--and that's why the second derivative test has the given result.

Fortunately, there's an easier way. Let $g(y)=y^2+\sin y$, determine its critical points and what type they are. From there, it follows what and which type of critical points $f$ has.

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Singularity theory is the branch of mathematics which deals with the classification of critical points. Recall that a critical point of a mapping $f : \mathbb{R}^m \to \mathbb{R}$ is a point $x_0 \in \mathbb{R}^m$ for which the Jacobian matrix $J_f$ has less than maximal rank at $x=x_0$, i.e. $\text{rank}(J_f)(x_0) =0.$

The most coarse classification is into non-degenerate and degenerate. A critical point is non-degenerate if the Hessian matrix is non-singular at $x=x_0$ and degenerate otherwise.

One way of distinguishing between the degenerate critical points is via their Milnor numbers $\mu(f) \in \mathbb{N}$, also known as the local multiplicity of the critical point. (The Milnor number is calculated as the dimension of a vector space formed by taking the quotient of the local ring of function germs by the Jacobian ideal generated by all partial derivatives.)

Non-degenerate critical points, also called Morse type critical points have $\mu(f) = 1.$ Degenerate critical points have $\mu(f) \ge 2.$ A function has an algebraically isolated (an isolated point even when extended over the complex numbers) critical point if and only if $\mu(f) < \infty.$ Critical points are put into equivalence classes, equivalent under diffeomorphic changes of variable in the source and the target. One property is that if $\mu(f) \neq \mu(g)$ then $f$ and $g$ are non-equivalent under $\mathscr{A}$-equivalence.

In the case of $f(x,y)=y^2 + \sin y$, the set of critical points is given by:

$$\{ (x,y) \in \mathbb{R}^2 : 2y + \cos y = 0 \}$$

which forms a curve in the $xy$-plane. This function has non-isolated critical points and so $\mu(f) = \infty$ for each and every critical point. This means that your critical points are very nasty, and does not belong to any of the well-known equivalence classes.

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