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How do you determine if #x/(x^2 -1)# is an even or odd function?

Answer 1

Ariana Alvarez

#x/(x^2-1)# is an odd function

An even function is one for which #f(-x) = f(x)# for all #x# in the domain.

An odd function is one for which #f(-x) = -f(x)# for all #x# in the domain.

In our example:

#f(x) = x/(x^2-1)#

#f(-x) = (-x)/((-x)^2-1) = (-x)/(x^2-1) = -x/(x^2-1) = -f(x)#

So #x/(x^2-1)# is an odd function.

graph{x/(x^2-1) [-10, 10, -5, 5]}

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Answer 2

Elijah Allen

To determine if the function ( \frac{x}{x^2 - 1} ) is even or odd, we evaluate ( f(-x) ) and compare it to ( f(x) ).

If ( f(-x) = f(x) ) for all ( x ) in the domain, the function is even.
If ( f(-x) = -f(x) ) for all ( x ) in the domain, the function is odd.
If neither condition 1 nor 2 holds, the function is neither even nor odd.

Evaluating ( f(-x) ): [ f(-x) = \frac{-x}{(-x)^2 - 1} = \frac{-x}{x^2 - 1} ]

Comparing ( f(-x) ) with ( f(x) ): [ \frac{-x}{x^2 - 1} \neq \frac{x}{x^2 - 1} ]

Since ( f(-x) ) is not equal to ( f(x) ) and also not equal to the negative of ( f(x) ), the function ( \frac{x}{x^2 - 1} ) is neither even nor odd.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.