Proving whether $f(x) =|x|^{1/2}$ is uniformly continuous for $f:\mathbb{R}\to\mathbb{R}$

Question

I need to prove following question: proving whether or not, $f(x) =|x|^{1/2}$ is uniformly continuous for $f:\mathbb{R}\to\mathbb{R}$

My attempt was based on this is not uniformly continuous. following definition and With taking $y=x/2$ at the result I could find some epsilon not providing the definition. I could prove that is continuous but I still suspect about .my attempt and I need hand.

Answer 1

We have the inequality that $\left||x|^{1/2}-|y|^{1/2}\right|\leq|x-y|^{1/2}$ because, say, $|x|\geq|y|$, then $|x|=|y+(x-y)|\leq|y|+|x-y|+2|y|^{1/2}|x-y|^{1/2}=\left(|y|^{1/2}+|x-y|^{1/2}\right)^{2}$, so $\left||x|^{1/2}-|y|^{1/2}\right|=|x|^{1/2}-|y|^{1/2}\leq|x-y|^{1/2}$.

Answer 2

Given $\epsilon >0$, there is $r>0$ such that $ |f(x)| < \epsilon$ for all $|x|<r$ there.

Outside that neighborhood of origin, say on $|x|>r/2$ the derivative is bounded, thus by mean value theorem uniformly continuous. Patching them together, we get uniform continuity of $f$ on all $\mathbb{R}$.