I recently became aware of the fact that lines of symmetry must pass through centers of mass, I was wondering if that could allow for a way to find the line of symmetry of functions as it removes the need to find another variable, namely the y-intercept, usually notated as $b$. Here is my work towards a solution, though it doesn't move very far, and more movement towards a solution or advice would be greatly appreciated!
Here is my work:
Asumming the function $f(x)$ has a line of symmetry of the form $y=mx+b$ then $\frac{2(mx+b)+f(x)(m^2+1)}{m^2+1}$ should be equal to $f(\frac{2m(f(x)-b)-x(m^2-1)}{m^2+1})$ for all $x$. Let's call the center of mass of $f(x)$ $(x_c,y_c)$. Assuming $f^{-1}(x)$ exists $y_c$ should equal $\lim_{a \rightarrow \infty} \frac{1}{2a} \int_{-c}^{c}f(x)dx$ and $x_c$ should equal $\lim_{a \rightarrow \infty} \frac{1}{2a} \int_{-c}^{c}f^{-1}(x)dx$. With this in mind, the line of symmetry can be rewritten as $y=m(x-x_c)+y_c$ and $b$ can be defined as $\lim_{a \rightarrow \infty}\frac{1}{2a} \int_{-c}^{c}f(x)-mf^{-1}(x)dx$.
That's as far as I can get, once I know this, I don't know how to move forward in hope of solving for $m$ and therefore finding the line of symmetry.