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How to prove that $I(0,-1)$ is the center of symmetry of the function
$$F(x)= x - \dfrac{2e^x}{(e^x -1)}$$ Is there any formula that I can directly apply?

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A general rule simply involves shifting the function. Proving that $(a,b)$ is a center of symmetry for $f(x)$ is the same as proving that the function

$$f(x+a)-b$$

is symmetric around the origin. This means you have to prove $$f(-x+a)-b=-(f(x+a)-b)$$

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Define $y(x) = F(x) + 1$. As y is shifted $1$ upwards its center should be $(0,0)$. Central symmetry in origin implies: $y(-x) = -y(x)$ You must demonstrate:

$F(-x) + 1 = -F(x) - 1$

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Thanks but I want a more genral rule that I can use for any point –  noname Nov 3 '13 at 13:17
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For any point $I(a,b)$, prove that
$$f(a-x)+ f(x) =2b$$ In your case: $a=0$ and $b=-1$.
Then you have to prove $f(-x)+f(x) =-2$

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