If the absolute value of a function is continuous, is the function continuous? If $|f(x)|$ is continuous at $a$, is $f(x)$ continuous at $a$?
I tried doing it using composite functions.
If $g(x)= |x|$, then $g\circ f(x)= |f(x)|$.
Since $g(x)$ and $g\circ f(x)$ are continuous, $f(x)$ is continuous.
I don't know if this is correct. Please help.
 A: Let $f(x)=-1$  if $x$ is rational, and let $f(x)=1$ if $x$ is irrational.
Or else more modestly let $f(x)=-1$ if $x\lt 17$, and $f(x)=1$ for $x\ge 17$. 
A: This is not always true. If $f(x)$ is a piecewise function such that
$f(x)=1$ for $x<a$ or $x=a$ and
$f(x)=-1$ for $x>a$
Then $|f(x)|$ is continuous at $a$ but $f(x)$ is not.
A: Maybe you are misunderstanding the logic behind it:
Let $g(x)$ be continuous, OK.
Now, saying

$g \circ f(x)$ is continous if $f(x)$ is continuous

doesn't imply that

$g \circ f(x)$ is not continuous if $f(x)$ is not continuous

Also, it's never true to say that $g \circ f(x)$ is continuous because $f(x)$ is continuous.

The single (and obvious) conclusion that we can take is that if $f(x)$ is continuous, then $g \circ f(x)$ certainly is continuous too. But if $f(x)$ is discontinuous, then $g \circ f(x)$ may be, or may be not.
So you can deduce nothing about $f(x)$ from the behaviour of $g \circ f(x)$.
A: In programming speak as I don't know how to express this in math speak:
f(x) = odd(x) ? x : -x;

Ed.: that would be
$$
f(x) = \begin{cases} x \quad \text{if $x$ is odd} \\ 
 -x \quad \text{otherwise} \end{cases}
$$
