Log is my first love

Algebra Level 3

Determine whether the function $f(x) = \ln(x+ \sqrt{1+x^2} )$ is an odd function or an even function.

There is insufficient information Neither even nor odd Even Function Odd Function

2 solutions

Josh Banister
Sep 5, 2015

$\begin{aligned}f(x) + f(-x) &= \ln(x+\sqrt{1 + x^2}) + \ln(-x+\sqrt{1 + (-x)^2}) \\ &= \ln \big((x+\sqrt{1 + x^2})(-x + \sqrt{1+x^2})\big) \\ &= \ln\big(-x^2 + (1+x^2)\big) \\ &= \ln(1) \\ &= 0 \end{aligned}$

Therefore $f(x) = -f(-x)$ and $f(x)$ is odd.

Sudeep Salgia
Sep 4, 2015

The way to check whether a function is odd or even is to check how is $f(-x)$ related to $f(x)$ . So,

$\displaystyle f(-x) = \ln ( -x + \sqrt{1 + x^2} ) = \ln \left( \frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} +x)}{( \sqrt{1 + x^2} +x)} \right)$ $\displaystyle \Rightarrow f(-x) = \ln \left( \frac{ 1+ x^2 - x^2 }{(\sqrt{1 + x^2} +x)} \right) = \ln \left( \frac{1}{\sqrt{1 + x^2} +x} \right)$
$\displaystyle \therefore f(-x) = - \ln \left( x + \sqrt{1 + x^2} \right) = - f(x)$

It means $f(x)$ is an odd function.

Moderator note:

Good explanation. It is not immediately obvious why one should attempt to rationalize the numerator. That follows from the observation that $( \sqrt{ 1+x^2 } -x ) ( \sqrt{ 1 + x^2 } + x ) = 1$ .

You miss one more step (although trivial): you need to show the function is not even. But a function that is even and odd must be identically zero, so just plug somewhere to make the function nonzero ( $x = 1$ , for example).

Ivan Koswara - 5 years, 9 months ago

I'd just like to add $f(x)=\sinh^{-1}(x)$ which is obviously odd.

Isaac Buckley - 5 years, 9 months ago

That's indeed a short method. But seeing the level of the problem I did not intend to use hyperbolic functions in the solution so that things remain simple.

Sudeep Salgia - 5 years, 9 months ago

f(1)=ln (1+root 2) f(-1)= ln( -1+root2 ) how its odd f(1)# -f(-1)

Patience Patience - 5 years ago

Log is my first love

2 solutions

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