How do you tell that the function f(x)=1/(x-1) is neither - that is aside from graphing it is?

Question

How do you tell that the function f(x)=1/(x-1) is neither - that is aside from graphing it.

By definition if f(-x) = f(x) then it is even
and if f(-x) is -f(x) then it is odd.

In this case a negative x will always generate a negative answer - therefore it should be odd... but the graph says otherwise.. Please explain without using a graph.

Answer 1

Did you mean for there to be an absolute value sign bracketing x-1? If not, you have the following:

Given: f(x) = 1/(x-1) (Note f(x) is undefined when x=1)
And by definition: if f(-x) = f(x) then it is even;
if f(-x) = -f(x) then it is odd.

First if f(-x) = f(x) then

1/(-x-1) = 1/(x-1) which implies

x-1 = -x-1 so x = -x, which is true only if x=0

So the function is only even when x =0

2nd, If f(-x) = -f(x) then

1/(-x-1) = - 1/(x-1) which implies

x-1 = -(-x-1) = x+1 which imples

-1 =1 which is never true.

So the function is never odd.

Answer 2

Your statement, “In this case a negative x will always generate a negative answer - therefore it should be odd...” is NOT what it means to be an odd function, in fact you already wrote what it means to be an odd function.

"f(-x) is -f(x) then it is odd”

That is, just because f(-x) is negative does not mean f(x) is odd, f(-x) = -f(x) means f(x) is odd

So if:

f(x) = 1/(x – 1)

f(-x) = 1/ (-x -1) = -1/(x+1) so since this does not “look” like -1/(x -1) it is not odd.

Also, as we can see, it is not even, so it is neither even or odd.

Answer 3

The function is not symmetric about x=0. Notice that even for some postive values of x (x<1) the value is negative. Nor is it symmetric about x=1, but it is postive for x>1, negative for x<1.

Answer 4

Necessary conditions (not complete)

odd: f(0) = 0
even: f(x) only has even power terms in x

Answer 5

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