# Strange Function

1. Jul 25, 2008

### roam

Here's a problem

Let k: R\{-1} → R be given by $$k(x) = \frac{2x-1}{x+1}$$

Prove that k is neither even nor odd.

That is strange!! But to prove it we go back to the definition;
A function, $$f: (-a,a) \rightarrow R$$ is said to be even if for all $$x \in (-a,a)$$ => $$f(x) = f(-x)$$
And it is odd if $$f(x) = -f(-x)$$.

So: if k is even then $$k(-x) = k(x)$$ for any x
and $$k(x) = -k(-x)$$ if it is odd.

I get the feeling that my proof doesn't seem right and somewhat unimplemented... (1) is there a better way to prove and conclude that k is neither even nor odd?
And (2) if it is neither even nor odd, then what could it be?

Last edited: Jul 25, 2008
2. Jul 25, 2008

### Defennder

Well, as you have stated it, all you need to do to prove that is to show that k(x) doesn't satisfy either the odd or even criterion. And there's nothing strange about a function being neither odd nor even since a whole lot of functions are neither odd nor even, such as e^x, ln(x) etc.

3. Jul 25, 2008

### HallsofIvy

I'm not sure why you think that function is "strange". It's a simple rational function. If it is because it is neither even nor odd, well most functions are neither even nor odd.

To be even, a function must satisfy, as you say, f(x)= f(-x) for all x.
To be odd, a function must satisfy, as you say, f(x)= -f(-x) for all x.

To prove something is NOT "true for all x", you only need to find a counterexample.

If f(x)= (2x-1)/(x+1), what is f(2)? What is f(-2)?

That's all you need.

4. Jul 28, 2008

### roam

Hall, that's because I thought since even functions are their own reflection in the y-axis. Even functions are symmetric about the y-axis.
Whereas other functions which are not even (odd functions) aren't like that.
Yeah, I understood the error I made.

Yes, I get it. Hi@Defennder

$$\frac{2 \times 2-1}{2+1}$$ = $$\frac{2 \times -2-1}{-2+1}$$
1 ≠ -5
But they don't equal, thus they don't satisfy neither of the conditions $$f(x) = f(-x)$$ or $$f(x) = -f(-x)$$.

^ is that a valid proof?

Last edited: Jul 28, 2008
5. Jul 28, 2008

### HallsofIvy

Yes, that is a perfectly valid proof: to show that a general statement is NOT true, it is sufficient to give a 'counter example': one example where it is not true.

Yes, it is true that an even function is symmetric about the x-axis. It is NOT true that if a function is not symmetric about the x-axis- that is that it is not even- it is necessarily "odd". An odd function is symmetric through the origin: if (x,y) is on the graph then extending a line from (x,y) through (0,0) an equal distance on the other side of (0,0), that is to (-x,-y), you are again on the graph: f(-x)= -f(x).

6. Jul 30, 2008

### vkroom

If one maps the excluded point (here -1) to the origin such the (new) domain is R\{0}, things fall back to the old definition of f(x) and f(-x). Which in this case would be $$\frac{2x-3}{x}$$. This is pretty simple now :)