# Odd or even function?

1. Aug 6, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
Determine whether the following function is odd or even or none:
$$f(x)=\log \left(x+\sqrt{1+x^2} \right)$$

2. Relevant equations

3. The attempt at a solution
For an even function, $f(-x)=f(x)$ and for an odd function $f(-x)=-f(x)$.
Replacing x with -x in the given function,
$$f(-x)=\log\left(-x+\sqrt{1+x^2}\right)$$
I don't see it being equal to f(x) or -f(x) so it should be neither odd nor even but the answer key states it is an odd function.

Any help is appreciated. Thanks!

2. Aug 6, 2013

### dirk_mec1

hint: simplify the argument.

3. Aug 6, 2013

### Pranav-Arora

The only thing I can think of is multiplying and dividing by $\sqrt{1+x^2}+x$, do you ask me this?

4. Aug 6, 2013

### LCKurtz

Why don't you just try it and see?

5. Aug 6, 2013

### Pranav-Arora

Woops, just tried that and it does come out to be -f(x). Thanks dirk_mec1! :tongue2:

6. Aug 6, 2013

### verty

I like this question, very nice it is.

Hint: compare f(x) and f(-x), what do you see inside the logs? They are begging to have something done to them...

You've solved it now I see, so I'll just mention it. They are conjugate expressions that deserve to be multiplied together.

7. Aug 6, 2013

### Pranav-Arora

Multiplied? Or should I add them?

8. Aug 6, 2013

### haruspex

Multiply the expressions inside the logs, which is the same as adding the logs.
Going back to the problem as given, the most natural thing to try is f(x)+f(-x). As soon as you see the form log(..)+log(..), the next step should be automatic.