Is f(x) uniformly continuous on [0, inf)?

Icebreaker
"Suppose f:[0, inf) -> R is such that f is uniformly continuous on [a, inf) for some a>0. Prove that f is uniformly continuous on [0, inf)."

But this is not true, is it? Consider the function

f(x)=\left\{\begin{array}{cc}x &\mbox{ if }x\geq 1\\ \frac{1}{x-1} &\mbox{ if }x<1\end{array}\right
 
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Hah! The question forgot to state that f is supposed to be a continuous map [0, +\infty) \rightarrow \mathbb{R}!
 
Phew! It's not just me then.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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