TITLE: show if function is even or odd
QUESTION [6 upvotes]: Suppose that we have equation:
$$f(x)=\frac{2^x+1}{2^x-1}$$
There is question if this function even or odd? I know definitions of even and odd functions, namely
even is if $f(-x)=f(x)$  and odd is if $f(-x)=-f(x)$ and when I put $-$ sign in function, found that this is neither even nor odd function, because $2^{-x}\ne-1 \times 2^x$, but my  book says that  it is even, so am I wrong? Please help me to clarify book is correct or me? Thanks

REPLY [12 votes]: Let's see what $f(-x)$ looks like:
$$
f(-x) = \frac{2^{-x} + 1}{2^{-x} - 1}
$$
Since $f(x)$ contains $2^x$ and not $2^{-x}$, let's multiply the numerator and denominator by $2^x$:
$$
f(-x) = \frac{2^x(2^{-x} + 1)}{2^x(2^{-x} - 1)} = \frac{1 + 2^x}{1-2^x} = - \frac{2^{x} + 1}{2^{x} - 1} = -f(x)
$$
This shows that the function is odd.

REPLY [7 votes]: The function is odd. You propably miss something in your calculation.
$ \displaystyle{ f(-x) = \frac{2^{-x}+1}{2^{-x}-1}= \frac{ \frac{1}{2^x} + 1 }{ \frac{1}{2^x} -1} = \frac{ \frac{1+2^x}{2^x}}{ \frac{1-2^x}{2^x}}  =-f(x)}$

REPLY [6 votes]: $$\frac{2^{-x}+1}{2^{-x}-1} = \frac{\frac 1 {2^x}+1}{\frac 1 {2^x}-1}$$
Now clear factions by multiplying numerator and denominator by $2^x$ and see what you have.

REPLY [4 votes]: We get that
$$
f(-x) = \frac{2^{-x}+1}{2^{-x}-1} = \frac{1+2^x}{1-2^x}= - \frac{2^x + 1}{2^x-1} = -f(x),
$$
so the function is odd.