# Even, odd or neither function

1. Sep 26, 2009

### skeeterrr

1. The problem statement, all variables and given/known data

Suppose g is a function defined for all x > 0 which satisfies the following properties for all a, b > 0:

g(1) = 0

g(a/b) = g(a) - g(b)

Determine whether the function f(x) = g(x+ root(x^2+1)) is even, odd or neither, and justify your answer.

2. Relevant equations

Odd function: f(-x) = -f(x)
Even function: f(-x) = f(x)

3. The attempt at a solution

Well I was thinking that g could be a logarithmic function, and x+ root(x^2+1) is neither even nor odd, so f(x) is a neither function...

I think I'm approaching this the wrong way... Can anyone help me out please?

2. Sep 26, 2009

### sylas

The insight that g is a logarithmic function is good.

Try solving f(x) + f(-x), and f(x) - f(-x). If f is odd or even, then one of these will have to reduce to zero.

3. Sep 28, 2009

### skeeterrr

Ok so I took time to examine this problem again.

g is a function defined for x>0, and satisfies the following properties for a,b > 0:

g(1) = 0, g(a/b) = g(a) - g(b)

Having seen that, I can assume that g is a logarithmic function because:

let h be log x

h is a function defined for x>0, and satisfies the following properties for a,b >0:

h(1) = log 1 = 0, h(a/b) = log (a/b) = h(a) - h(b) = log a - log b

So now I can assume g is a logarithmic function.

Now I have to show if f(x) = g(x+ root((x^2)+1)) is even, odd, or neither.

I'll check if f(x) is odd or not:

f(-x) = f(x) <- definition of an odd function

g(-x+root(((-x)^2)+1) = -g(x+root((x^2)+1))

since g is a logarithmic function (from the assumption above, I can replace g with log(base 10))

log(-x+root(((-x)^2)+1) = -log(x+root((x^2)+1))

log(-x+root((x^2)+1) = -log(x+root((x^2)+1))

the left hand side can not be explicitly interpreted whether it equals to -f(x) or not, so I have to rationalize the left hand side

log[(-x+root((x^2)+1)(x+root((x^2)+1/x+root((x^2)+1)] = -log(x+root((x^2)+1))

log(-x^2+x^2+1/x+root(x^2)+1)) = -log(x+root((x^2)+1))

log(1/x+root(x^2+1) = -log(x+root((x^2)+1))

from the algebraic properties of logarithms (log(x/y)=logx-logy), I can make the left hand side equal to:

log 1 - log(x+root(x^2)+1)) = -log(x+root((x^2)+1))

since g(1) = 0, log (1) = 0,

0 - log(x+root(x^2)+1)) = -log(x+root((x^2)+1))

- log(x+root(x^2)+1)) = -log(x+root((x^2)+1))

now i replace log with g

-g(x+root(x^2)+1)) = -g(x+root((x^2)+1))

g(-x+root((-x)^2)+1)) = -g(x+root((x^2)+1)) (from above)

therefore f(-x) = -f(x)

so f(x) is an odd function

Phew...

Now can anyone confirm this please? ;) I apologize if this is hard to read, I do not know how to use latex coding...

Last edited: Sep 28, 2009
4. Sep 29, 2009

### thofer

Your solution (f(-x)=-f(x)) is correct. Just two remarks:
1. You do not need to say that the function g is a logarithm (and it is not forcibly of base 10).
2. You can proceed as you did to find the answer, that is, you start with f(-x)=-f(x) and try to reduce. But if you want to write it properly once you know it's true you have to start with f(-x) and you should arrive at -f(x).
So my proposition would be:
Set
$$f(x) = g(x+\sqrt{x^2+1}).$$
We have
$$f(-x) = g(-x+\sqrt{(-x)^2+1}) = g(-x+\sqrt{x^2+1}) = g\left(\frac{-x^2+x^2+1}{x+\sqrt{x^2+1}}\right) = g(1)-g(x+\sqrt{x^2+1}) = -f(x),$$
thus, $$f$$ is an odd function.
If $$g$$ is zero everywhere $$f$$ is also an even function.

5. Sep 29, 2009

### skeeterrr

Yes, moments after I posted this, I realized I did not have to replace g with a logarithm. Thanks for your input!