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Can there be a function $f$ for which $f(x) = f(|x|)$ (excluding $f(x) = x^2$ or any even exponent)?

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    $\cos{x}$? Or let $g$ be any function, and consider $g(x)+g(-x)$. – 2017-02-15
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    Hehe, $|x|$ ! (and from there, all $f(|x|)$). – 2017-02-15

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Yes any even function will satisfy the given condition f(x) = f(|x|) As it is standard definition of even function e.g. y = |x| , y = cos(x) Actually any function giving same value of y for -ve x and +ve x where x will have same magnitude in both cases

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    I was going to upvote, but couldn't after I saw "+ve" :) – 2017-02-15
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    We generally not show +ve sign but actually it is +ve x – 2017-02-15
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    I just meant "+ve" as an abbreviation; it drives me crazy for some reason. – 2017-02-15
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    Great.......... – 2017-02-15
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    So "$+$" stands for "positi", and "$-$" stands for "negati"? – 2017-02-15
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    Yes you are right #Jonas – 2017-02-15
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Of course. Take any function defined on the positive reals and zero. Then define $f(-x)=f(x)$. My favorite is $f(x)=\text{cheese}$ for any $x \in \Bbb R$

This is very flip, but there is a real point here. There are lots of functions out there, most of which do not have any nice description at all. Most are not continuous, let alone differentiable, but the ones we draw are (almost) always differentiable. You did not specify the range, so I could define a function from $\Bbb R$ to $\{\text{cheese}\}$. If you had demanded $\Bbb R \to \Bbb R$ I could have said $f(x)=18000$ instead. Just within functions from $\Bbb R \to \Bbb R$ I have shown how to get $2^{\mathfrak c}$ of them.

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    Do the set of functions and the set of even functions have the same cardinality ? – 2017-02-15
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    @YvesDaoust: Yes, they are both $\mathfrak c^{\mathfrak c}$. There are $\mathfrak c$ positive numbers and each one gets $\mathfrak c$ choices of the function value. – 2017-02-15
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    Agreed, but can we establish an easy bijection between these ? – 2017-02-15
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    @YvesDaoust: start with your favorite bijection $\Bbb R \leftrightarrow \Bbb R^+$ Then each function of $\Bbb R$ matches a function on $\Bbb R^+$. Extend the latter by symmetry. – 2017-02-15
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All functions such that $f(x)=f(|x|)$ can be written as $g(|x|)$, where $g$ coincides with $f$ in the non-negative domain.

Conversely, for any function $f$, $f(|x|)$ has the desired property.

Hence,

$$5,\\|x|,\\\lfloor|x|\rfloor,\\ |x|^k=|x^k|,\\\cos(|x|)=\cos x,\\\sin(|x|)=|\sin x|,\\\ln|x|,\\a|x|^2+b|x|+c=ax^2+b|x|+c\\\cdots$$