Finding the center of function in general Let $f(x)$ be a function . How we can find the center of $f(x)$ (i.e. center of symmetry ) if it exist ? (And also prove it doesn't have center . )
For example consider $\log_{10} \frac{x-2}{x-1}$ . If we graph the function , it seems $\omega(1.5 , 0)$ is the center but I don't know how to show it .
 A: A function is symmetric with respect to $\,(x_0,y_0)\,$ iff $\,f(x_0-x)+f(x_0+x) = 2 y_0\,$. In this case:
$$
\begin{align}
2y_0 &= \log\left(\frac{x_0-x-2}{x_0-x-1}\right)+\log\left(\frac{x_0+x-2}{x_0+x-1}\right) \\[5px]
 &= \log\left(\frac{(x_0-x-2)(x_0+x-2)}{(x_0-x-1)(x_0+x-1)}\right) \\[5px]
 &= \log\left(\frac{(x_0-2)^2-x^2}{(x_0-1)^2-x^2}\right)
\end{align}
$$
It follows that:
$$
\begin{align}
\frac{(x_0-2)^2-x^2}{(x_0-1)^2-x^2} = 10^{2y_0} \;\;\iff\;\; (x_0-2)^2-x^2 = 10^{2y_0}\big((x_0-1)^2-x^2\big)
\end{align}
$$
For the equality to hold for all $\,x\,$, the coefficients of $\,x^2\,$ must be equal between the two sides, so $\,10^{2y_0}=1 \iff y_0=0\,$. This leaves the equation $\,(x_0-2)^2=(x_0-1)^2 \iff x_0=\dfrac{3}{2}\,$, so in the end the center of symmetry is $\,(x_0,y_0)=\left(\dfrac{3}{2}, 0\right)\,$.

[ EDIT ]  As noted in a comment, if the center of symmetry is the origin $\,(x_0,y_0)\,$ the condition of symmetry reduces to $\,f(-x)+f(x) = 0 \iff f(-x)=-f(x)\,$. The latter means that $\,f(x)\,$ is an odd function, which is indeed equivalent to central symmetry across the origin.

In the general case, the condition of symmetry $\,f(x_0-x)+f(x_0+x) = 2 y_0\,$ can also be written as $\,g(-x)+g(x)=0\,$ where $\,g(x)=f(x+x_0)-y_0\;$ i.e. $\;f(x)\,$ has a center of symmetry at $\,(x_0,y_0)\,$ iff the translation of its graph that maps $\,(x_0,y_0) \mapsto (0,0)\,$ represents an odd function.
