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In homework there is such problem:

Express $\;f(x)=\dfrac{x − 1}{x + 1}\;$ as the sum of an even and an odd function.
(Simplify as much as possible.)

This function is not even and neither odd. Also if we take it as division of 2 functions, neither $x - 1$ nor $x + 1$ are odd or even... so I'm confused...

amWhy
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Tsiskreli
  • 63

2 Answers2

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Define

  • $f_e(x) = \frac12\left(f(x)+f(-x)\right)$
  • $f_o(x) = \frac12\left(f(x)-f(-x)\right)$

Then $f_e$ is even and $f_o$ is odd and $f_e+f_o=f$ for any given $f$

In your special case $\displaystyle f(x) = \frac{x-1}{x+1}$, so $$f_e(x) = \frac12\left(\frac{x-1}{x+1} + \frac{-x-1}{-x+1} \right) = \ldots$$ $$f_o(x) = \frac12\left(\frac{x-1}{x+1} - \frac{-x-1}{-x+1} \right) = \ldots$$ you just have to simplify.

user127.0.0.1
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4

I know it's late, but you could also simply multiply both numerator and denominator by $(x-1)$. $$ \begin{align} \frac{x-1}{x+1} &= \frac{x-1}{x+1} \cdot \frac{x-1}{x-1} \\ &= \frac{x^2-2x+1}{x^2-1} \\ &= \frac{x^2+1}{x^2-1} + \frac{2x}{1-x^2} \\ \end{align} $$

The left-hand term is even since it is of the form $even/even$, and the right-hand term is odd since it has the form $odd/even$.

henriq
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