Can't determine even function

Question

Given $f(x)= \cos\left(\log( x+\sqrt{1+x^2}\right))$

As to check if even or not, we usually use $f(-x)=f(x)$

But here if we do that the even nor odd However when graph is plotted we get even. How do we determine such function as even without plotting graphs...

Answer 1

It is indeed an even function; you should plug $-x$ again. You might also investigate the relationship between $(-x + \sqrt{1 + x^2)}$ and $(x + \sqrt{1 + x^2})$ (Hint: What happens when you multiply them? What does that say about their logarithms?) Finally, just note that cosine is even function. (I cannot comment right now, so if someone wants to move this to comments, be my guest)

Answer 2

As the cosine is an even function, its argument has to be an even or an odd function.

Then the argument of the logarithm must be either an even function, or one such that $$f(-x)=\frac1{f(x)}.$$

Taking $x=1$, we see that $-1+\sqrt2\ne1+\sqrt2$ and evenness is rejected. Now, as

$$\left(-x+\sqrt{1+(-x)^2}\right)\left(x+\sqrt{1+x^2}\right)=1+x^2-x^2,$$ the argument of the logarithm indeed has this "inverse" property.

To summarize,

$$f(-x)=\frac1{f(x)},\\\log(-f(x))=-\log(f(x)),\\\cos(\log(f(-x))=\cos(\log(f(x)).$$