<s>[INST] How to show an exponential function is symmetric w.r.t. the y-axis.? The function $\dfrac{1}{2}x + \dfrac{x}{e^x -1}$ is symmetric w.r.t. the y-axis, and I want to demonstrate this. So I basically have to show that $$\dfrac{1}{2}x + \dfrac{x}{e^x -1} = - \dfrac{1}{2}x - \dfrac{x}{e^{-x} -1}$$
But I can't do this algebraically. Can someone help me with this?
 [/INST] Here is a way to algebraically get the answer from where you left off.
$ - \frac{1}{2}x \ - \ \frac{x}{e^{-x} \ - \ 1} \ = \ - \frac{1}{2}x \ - \ \frac{xe^x}{1 \ - \ e^x } \ = \ - \frac{1}{2}x \ + \ \frac{xe^x}{e^x \ - \ 1} $
Now , long division will give ,
$ \frac{xe^x}{e^x \ - \ 1} \ = \ x \ + \ \frac{x}{e^x \ - \ 1} $
And substituting gives , 
$ - \frac{1}{2}x \ + \ x \ + \ \frac{x}{e^x \ - \ 1} \ = \ \frac{1}{2}x \ + \ \frac{x}{e^x \ - \ 1} $
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