How to show that if $f^2(x)$ is uniformly continous function then f is uniformly continous $f:R\to [0,\infty)$ is function such that $f^2(x)$ is uniformly continuous on R then I have to show that f is uniformly continuous ?
My attempt :
$|f^2(x)-f^2(y)|<\epsilon$ for  $|x-y|<\delta$
then 
$|f(x)-f(y)<\epsilon/|f(x)+f(y)|$ for  $|x-y|<\delta$
My problem is that how to control above difference as f may be 0 at both x and y
SO how to show above is uniformly continuous 
Any help will be appreciated

I think the result holds for all $f\in C(\mathbb{R})$. You could use mean value theorem: if $f(x)f(y)<0$ then there is some $z$ between $x$ and $y$ such that $f(z)=0.$ The details is omitted. 
