= Solution
{c}
For any $x,y\in\mathbb R$, rationalizing the difference of <square roots> gives
$$
|g(x)-g(y)|=\frac{|f(x)-f(y)|}{\sqrt{f(x)}+\sqrt{f(y)}}\leq\frac12|f(x)-f(y)|,
$$
since $f(x),f(y)\geq1$. Given $\varepsilon>0$, use <uniform continuity> of $f$ with tolerance $2\varepsilon$ to obtain $\delta>0$. Whenever $|x-y|<\delta$, the displayed estimate yields $|g(x)-g(y)|<\varepsilon$. Thus \b[$g$ has <uniform continuity> on $\mathbb R$.] The <square root> <function> is in fact <Lipschitz continuous> with constant $1/2$ on $[1,\infty)$.
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