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?
 A: $$f(x) =\dfrac{1}{2}x + \dfrac{x}{e^x -1} = \dfrac {x(e^x - 1) + 2x}{2(e^x - 1)} = \dfrac{xe^x - x + 2x}{2(e^x - 1)} = \dfrac{xe^x + x}{2(e^x - 1)} = \dfrac{x(e^x + 1)}{2(e^x - 1)}$$
At this point, you should be able to convince yourself that $f(x) = f(-x)$.
A: $$f(x)=\dfrac{1}{2}x + \dfrac{x}{e^x -1}$$
$$=\frac{x(e^x+1)}{2(e^x-1)}$$ now what do you get for $f(-x)$. its the same $f(x)$.
A: This means you have to show the function is even. Write $f(x)=\frac{x}{2}+ \frac{x}{e^x-1}$, which can be rewritten as $\frac12 \frac{x(e^x-1)+2x}{e^x-1}=\frac12 \frac{xe^x+x}{e^x-1}$. Then $f(-x)=\frac12 \frac{-xe^{-x}-x}{e^{-x}-1}\cdot \frac{-e^{-x}}{-e^{-x}}=\frac12 \frac{x +xe^x}{-1+e^x}=f(x)$ as desired.
A: 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} $
