How do you know if $h (x) = \frac{x}{x^{2} - 1}$ $h(x) = x / (x^2 - 1)$ is an even or odd function?

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Answer 1 · 2015-12-31T03:18:33

Examine $h (- x)$ $h(-x)$ to see that $h (x)$ $h(x)$ is an odd function.

A function $f (x)$ $f(x)$ is even if $f (- x) = f (x)$ $f(-x) = f(x)$ .
A function $f (x)$ $f(x)$ is odd if $f (- x) = - f (x)$ $f(-x) = -f(x)$

Then, to see if $h (x)$ $h(x)$ is even or odd, we examine $h (- x)$ $h(-x)$ .

$h (- x) = \frac{- x}{{(- x)}^{2} - 1}$ $h(-x) = (-x)/((-x)^2-1)$

$= - \frac{x}{x^{2} - 1}$ $= -x/(x^2-1)$

$= - h (x)$ $= -h(x)$

Thus $h (x)$ $h(x)$ is odd.

How do you know if h(x) = x / (x^2 - 1)h(x)=xx2−1h(x) = x / (x^2 - 1) is an even or odd function?