Find critical points of parametric function
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Hi, I want to find the critical points and the value of the function at those critical points, of the following function:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1643941/image.png)
where
are positive constants. I tried using this code:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1643946/image.png)
clear; clc; close all;
syms y w ws R L C;
y = ((w+ws)*(w^2+ws^2)*(L^2*C^3))/((w*R^2*C+(w^2*L*C-1)^2)*(ws*R^2*C+(ws^2*L*C-1)^2));
dy = diff(y,w);
critical_points = solve(dy == 0);
The critical points calculated:
root(C^2*L^2*z^6 + 2*C^2*L^2*ws*z^5 + 3*C^2*L^2*ws^2*z^4 + 2*C*L*z^4 + 4*C^2*L^2*ws^3*z^3 - 2*C*R^2*z^3 - C*R^2*ws*z^2 - 2*C*L*ws^2*z^2 - 3*z^2 - 4*C*L*ws^3*z - 2*ws*z + C*R^2*ws^3 - ws^2, z, 1)
root(C^2*L^2*z^6 + 2*C^2*L^2*ws*z^5 + 3*C^2*L^2*ws^2*z^4 + 2*C*L*z^4 + 4*C^2*L^2*ws^3*z^3 - 2*C*R^2*z^3 - C*R^2*ws*z^2 - 2*C*L*ws^2*z^2 - 3*z^2 - 4*C*L*ws^3*z - 2*ws*z + C*R^2*ws^3 - ws^2, z, 2)
root(C^2*L^2*z^6 + 2*C^2*L^2*ws*z^5 + 3*C^2*L^2*ws^2*z^4 + 2*C*L*z^4 + 4*C^2*L^2*ws^3*z^3 - 2*C*R^2*z^3 - C*R^2*ws*z^2 - 2*C*L*ws^2*z^2 - 3*z^2 - 4*C*L*ws^3*z - 2*ws*z + C*R^2*ws^3 - ws^2, z, 3)
root(C^2*L^2*z^6 + 2*C^2*L^2*ws*z^5 + 3*C^2*L^2*ws^2*z^4 + 2*C*L*z^4 + 4*C^2*L^2*ws^3*z^3 - 2*C*R^2*z^3 - C*R^2*ws*z^2 - 2*C*L*ws^2*z^2 - 3*z^2 - 4*C*L*ws^3*z - 2*ws*z + C*R^2*ws^3 - ws^2, z, 4)
root(C^2*L^2*z^6 + 2*C^2*L^2*ws*z^5 + 3*C^2*L^2*ws^2*z^4 + 2*C*L*z^4 + 4*C^2*L^2*ws^3*z^3 - 2*C*R^2*z^3 - C*R^2*ws*z^2 - 2*C*L*ws^2*z^2 - 3*z^2 - 4*C*L*ws^3*z - 2*ws*z + C*R^2*ws^3 - ws^2, z, 5)
root(C^2*L^2*z^6 + 2*C^2*L^2*ws*z^5 + 3*C^2*L^2*ws^2*z^4 + 2*C*L*z^4 + 4*C^2*L^2*ws^3*z^3 - 2*C*R^2*z^3 - C*R^2*ws*z^2 - 2*C*L*ws^2*z^2 - 3*z^2 - 4*C*L*ws^3*z - 2*ws*z + C*R^2*ws^3 - ws^2, z, 6)
I'm not sure how to interpret the results in critical_points. What is z? and why are they all the same? what did I miss?
Another this, is there any easy way to place them back into the function to calculate its value at those points? couldn't find anything useful in the documentation, but I'm still new, so probably just missed it.
Thank you so much for your time and attention!
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