6
$\begingroup$

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

  • 0
    You can always [graph the function](http://www.wolframalpha.com/input/?i=%282%5Ex%2B1%29%2F%282%5Ex-1%29) and see that it's definitely not even, that it in fact looks like it's odd (something others have proved below).2012-06-17

4 Answers 4

12

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.

7

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)}$

  • 0
    yes i was thinking,that neither even or odd is this function,so have not finished calculation2012-06-17
6

$\frac{2^{-x}+1}{2^{-x}-1} = \frac{\frac 1 {2^x}+1}{\frac 1 {2^x}-1}$

Now clear factions by multiplying numerator and denominator by $2^x$ and see what you have.

4

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.