Determine whether a logarithmic function is odd or even

[YoungPhysicist]

Homework Statement

Determine function
$$f(x) = \log(\sqrt{x^2+1}+x)$$
Is odd or even.

Homework Equations

$\log(a+b) = \log(a) + \log(1+b/a)$

The Attempt at a Solution

First I thought it is a even function without considering the x at the end, which of course isn't the actual case.
Then I tried to use the formula above but still failed to determine it.
Maybe the formula works just that I can't see the point.
Are there any insights to how should I determine it?I am new to functions properties. Thanks!

Edit:I plot the function in a function drawer and found it to be an odd one, which cause I still don't know how to do it, the question remains.

 

[member 587159]

When I looked at the function, I immediately thought it was neither odd nor even. I then graphed the function and it looked pretty "odd". So I was mistaken!

We have to prove:

$f(-x) = -f(x)$

I suggest you start with the right side and use the rule $-\log(x) = \log\frac{1}{x}$. You will then have a square root in the denominator, and there is a known trick to remove roots from the denominators (the same trick you use when you solve limits with roots).

Let me know if my hints were clear enough to get the right answer.

 

[SammyS]

Homework Statement

Determine function
$$f(x) = \log(\sqrt{x^2+1}+x)$$
Is odd or even.

Homework Equations

$\log(a+b) = \log(a) + \log(1+b/a)$

The Attempt at a Solution

First I thought it is a even function without considering the x at the end, which of course isn't the actual case.
Then I tried to use the formula above but still failed to determine it.
Maybe the formula works just that I can't see the point.
Are there any insights to how should I determine it?I am new to functions properties. Thanks!

Edit:I plot the function in a function drawer and found it to be an odd one, which cause I still don't know how to do it, the question remains.

I like to use a test similar to$,\ -f(x) = f(-x) \,, \$ which was suggested by @Math_QED .

That is: See if you can show that $\ -f(-x)$ can be manipulated to give you $f(x)$ .

 

[YoungPhysicist]

This is what I get so far:
Since $-\log(x) = \log(1/x)$, I have to prove:
$$\log(\sqrt{x^2+1}-x) = \log(\frac{1}{\sqrt{x^2+1}+x})$$
Which I lack the

there is a known trick to remove roots from the denominators (the same trick you use when you solve limits with roots).

Sorry @Math_QED ,but what is the trick? I only cover really really basic rationalizing denominators in junior high so I don’t know how to do that

 

[SammyS]

This is what I get so far:
Since $-\log(x) = \log(1/x)$, I have to prove:
$$\log(\sqrt{x^2+1}-x) = \log(\frac{1}{\sqrt{x^2+1}+x})$$
Which I lack the

Sorry @Math_QED ,but what is the trick? I only cover really really basic rationalizing denominators in junior high so I don’t know how to do that

Well that's exactly what you need to do. Either rationalize the denominator on the right hand side of the equation, or rationalize the numerator for the expression on the left hand side.

 

[YoungPhysicist]

Well that's exactly what you need to do. Either rationalize the denominator on the right hand side of the equation, or rationalize the numerator for the expression on the left hand side.

Sorry, But my problem is that I don’t know the method to rationalize the denominator of
$$ \frac{1}{\sqrt{x^2+1}+x}$$it’s I don’t know how to do it,i.e, I don’t know multiplying it by what.

 

[Delta2]

Your denominator is of the form $a+b$ with $a=\sqrt{x^2+1},b=x$. If you multiply it by $a-b$ then you ll get $a^2-b^2$ and it will eventually get rationalized.
So multiply both numerator and denominator by $a-b=\sqrt{x^2+1}-x$ and see what you get.

 

[YoungPhysicist]

Thanks @Delta2 I remembered that formula
I just kept think about $(a+b)^2 = a^2+2ab+b^2$ and can’t get rid of the square rootSo the solution goes like this:

$$ (\frac{1}{\sqrt{x^2+1}+x}) \cdot (\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x})\\

=\frac{\sqrt{x^2+1}-x}{x^2+1-x^2}\\

=\sqrt{x^2+1}-x

$$

Which solves the problem.
Thanks again for @SammyS ,@Math_QED and @Delta2 for helping me out!

 

[SammyS]

Thanks @Delta2 I remembered that formula
I just kept think about $(a+b)^2 = a^2+2ab+b^2$ and can’t get rid of the square root So the solution goes like this:
$$ (\frac{1}{\sqrt{x^2+1}+x}) \cdot (\sqrt{x^2+1}-x)\\

=\frac{\sqrt{x^2+1}-x}{x^2+1-x^2}\\

=\sqrt{x^2+1}-x

$$
Which solves the problem.
Thanks again for @SammyS ,@Math_QED and @Delta2 for helping me out!

That should begin as:

$\displaystyle \left(\frac{1}{\sqrt{x^2+1}+x} \right) \cdot \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x} \right)$​

.

 

[YoungPhysicist]

That should begin as:

$\displaystyle \left(\frac{1}{\sqrt{x^2+1}+x} \right) \cdot \left( \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}-x} \right)$​

.

Oh yeah,right. Thread Fixed.

 

[ehild]

Just a note: f(x) is the inverse hyperbolic sine function, which is odd. http://www.sosmath.com/trig/hyper/hyper03/hyper03.html

 

[robphy]

Note: for odd $f$ , the relation $f(x)=-f(-x)$ is equivalent to $f(x)+f(-x)=0$.

So,
$\begin{align*} 0 &\stackrel{?}{=}\log(\sqrt{x^2+1}+x)+\log(\sqrt{(-x)^2+1}+(-x))\\ &\stackrel{?}{=}\log( \sqrt{x^2+1}+x)+\log(\sqrt{x^2+1}-x)\\ &\stackrel{?}{=}\log(( \sqrt{x^2+1}+x)(\sqrt{x^2+1}-x))\\ &\stackrel{\checkmark}{=}\log(\qquad \qquad \qquad1\ \qquad\qquad\qquad )\\ \end{align*}$