Is the function $f (x) = \frac{x^{2}}{x - 1}$ $f(x)=x^2/(x-1)$ even, odd or neither?

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Is the function $f (x) = \frac{x^{2}}{x - 1}$ $f(x)=x^2/(x-1)$ even, odd or neither?

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Answer 1 · 2015-09-01T14:09:41

It is neither odd or even.

Explanation:

First of all if you want to discuss if a function is odd or even, the domain of the function must be symetrical to 0 (in other words if $x \in D$ $x in D$ then $- x \in D$ $-x in D$ ).

Function given in this task does not satisfy this condition, because $- 1 \in D$ $-1 in D$ but $1 cancel(in) D$ .

Answer 2 · 2015-09-01T14:15:56

$f$ is neither even nor odd.

Explanation:

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

$= x^2/(-x-1) = x^2/-(x+1) = -x^2/(x+1)$

$f(-x) != f(x)$ (that is $f(-x)$ is not always the same as $f(x)$ ), so $f$ is not even

$f(-x) != -f(x)$ (that is $f(-x)$ is not always the same as $-f(x)$ , so $f$ is not odd.

$f$ is neither even nor odd.

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Is the function $f (x) = \frac{x^{2}}{x - 1}$ $f(x)=x^2/(x-1)$ even, odd or neither?

2 Answers

Explanation:

Explanation:

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Is the function f(x)=x^2/(x-1)f(x)=x2x−1f(x)=x^2/(x-1) even, odd or neither?

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Is the function $f (x) = \frac{x^{2}}{x - 1}$ $f(x)=x^2/(x-1)$ even, odd or neither?