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Visualization of a numerical method for determining the type of extrema of functions with two variables on a contour map without using derivatives

For the function f(x,y) with two variables on a contour map, a numerical method is proposed to determine the nature of the extrema without using the derivatives. The analysis is based on the composite function Δf(α) - changes of the function f(x,y) for the corresponding points on the circles (r; α) and (r+Δr; α). The applet determines areas of monotonic increase or decrease for the analyzed function f(x,y) on a test circle that is described around a critical point on the contour map. ✱If Δf(α)<0 for α∈[0,2 π ], then f(x,y) decreases at the ends of the radii r and r+Δr of the circles (for each α), i.e. there is a local maximum at their center. ✱If Δf(α)>0 for α∈[0,2 π ], then f(x,y) increases at the ends of the radii r and r+Δr of the circles (for each α), i.e. there is a local minimum at their center. ✱If Δf(α) for α∈[0,2 π ], is an alternating function with zeros (in the color "Deep Sky Blue"), then f(x,y) has increasing and decreasing function intercepts at the edges of these circles, i.e. there is a saddle point in the center of these circles. Near this point, the surface has the shape of a saddle around the critical point: - concave upwards in one direction, - concave downwards in another direction. *The applet offers the possibility to check the accuracy of these calculations by tracing (button - "Trace On") the calculated monotonicity limits. **In case index=1 it is possible to enter functions from the input box. ***Further improvements of the described method can be found in the applet.
A detailed solution of the problem for the case index=10 according to the above diagram is discussed in detail in the Calculus "What are the extrema and saddle points of f(x,y)=xy(1-x-y)?

Function graphs (index=3) with the same contour plot: f(x,y)=(x²-y²)ᴷ for k=1 and k=2

Function graphs (index=3) with the same contour plot: f(x,y)=(x²-y²)ᴷ for k=1 and k=2
1. Concentric closed contour lines always indicate either a local minimum or a local maximum. Critical points that are not a local maximum or minimum are usually saddle points. It is a crossing of two contour lines of f(x,y). The surface around the critical point has the shape of a saddle: - concave upwards in one direction, - concave downwards in another. 2. If a contour line intersects itself, the point could be a ★ Saddle point, ★ local minimum, or ★ local maximum. Here is a pair of function graphs with the same contour plot. The figure considers two cases for index=3: (a) for k=1 -Hyperbolic paraboloid: z=x2-y2 and (b) for k=2 -z=(x2-y2)2. For a discussion, see the article "How to read contour plot?". A similar case occurs for index=1: z=x y and z=(x y)2. You can see this case for yourself by setting the saddle point to a local maximum.

Example (index=7): f(x,y)=x⁶+y⁶-15(x²+y²)

Example (index=7): f(x,y)=x⁶+y⁶-15(x²+y²)
In this example, there is one local maximum point, three local minimum points, and three saddle points.

Example (index=8): f(x,y)=3(1-x²) exp(-(x-0.5)²-(y+1)²)-2(0.2x-x³-y⁵)exp(-x²-y²)sin(x-y)

Example (index=8): f(x,y)=3(1-x²) exp(-(x-0.5)²-(y+1)²)-2(0.2x-x³-y⁵)exp(-x²-y²)sin(x-y)

Example (index=9): Is N a saddle point?

Example (index=9): Is N a saddle point?
Is N a saddle point here? https://www.wolframalpha.com/input?i=%28x%5E2%2B+y%5E2%29%5E2%2B3+x%5E2+y-+y%5E3+ f(x, y) = (x² + y²)² + 3x² y - y³; fx (x,y)=4x³+2xy(2y+3); fy(x,y)=x²(4y+3)+y²(4y-3); fxx(x,y)=12x² + 4y² + 6 y; fxy(x,y)=8 xy+6 x; fyx(x,y)=8 xy+6 x; fyy(x,y)=4x²+12y²-6y; D(x,y)=|fxx*fyy-fxy*fyx|; D(x,y)= (12x² + 4y² + 6y) (4x² + 12y² - 6y) - (8x y + 6x)² = =12 (4x⁴ + 8x² y² - 12x² y - 3x² + 4y⁴ + 4y³ - 3y²); ⇒fx(0,0)=0; fy(0,0)=0 ; D(0,0)=0 Regarding the case D=0 in the literature we have ambiguous arguments: https://en.wikipedia.org/wiki/Second_partial_derivative_test ... critical point (0, 0) the second derivative test is insufficient, and one must use higher order tests or other tools to determine the behavior of the function at this point. (In fact, one can show that f takes both positive and negative values in small neighborhoods around (0, 0) and so this point is a saddle point of f.)

Scheme for the calculation of the stationary points of the 2-variable function

Scheme for the calculation of the stationary points of the 2-variable function
From https://ayraethazide.tumblr.com/post/74061500099/here-are-some-notes-on-the-classification-of A detailed solution of the problem for the case index=10 according to the above diagram is discussed in detail in the Calculus "What are the extrema and saddle points of f(x,y)=xy(1-x-y)?