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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

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    $\begingroup$ As this example shows, in order to show that a function is neither even or odd, it is not enough simply to invert the variable and get something that looks different. The only reliable way to conclude neither-even-nor-odd is to find a concrete number such that you can plug it and its negative into the function and get results that are neither equal or opposite. $\endgroup$ – Henning Makholm Jun 17 '12 at 14:18
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    $\begingroup$ This example also shows that, sometimes, books can be wrong! This is not an even function, as almost every answer below states explicitly. $\endgroup$ – Cameron Buie Jun 17 '12 at 15:43
  • $\begingroup$ You can always graph the function and see that it's definitely not even, that it in fact looks like it's odd (something others have proved below). $\endgroup$ – Ken Bloom Jun 17 '12 at 17:02
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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.

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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)}$

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  • $\begingroup$ yes i was thinking,that neither even or odd is this function,so have not finished calculation $\endgroup$ – dato datuashvili Jun 17 '12 at 13:47
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$$\frac{2^{-x}+1}{2^{-x}-1} = \frac{\frac 1 {2^x}+1}{\frac 1 {2^x}-1}$$

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

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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.

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