= Solution
A <function> $f:I\to\mathbb R$ is <uniformly continuous> if
$$
\forall\varepsilon>0\ \exists\delta>0\ \forall x,y\in I:\quad |x-y|<\delta\Longrightarrow|f(x)-f(y)|<\varepsilon.
$$
The crucial point is that $\delta$ works throughout the <interval>, not just near one chosen point.
\b[Yes, it is bounded on the closed unit interval.] Choose $\delta$ for $\varepsilon=1$, and choose finitely many grid points $t_j$ such that every $x\in[0,1]$ is within $\delta$ of some $t_j$. Then $|f(x)|\leq1+\max_j|f(t_j)|$, a finite bound. Equivalently, <uniform continuity> implies <continuity>, and a continuous real <function> on a <compact set> is bounded.
\b[Yes, a bounded <derivative> gives <uniform continuity> on the half-line.] If $|f'|\leq K$, the <mean value theorem> gives $|f(x)-f(y)|\leq K|x-y|$. This is a <Lipschitz condition>, so for $K>0$ use $\delta=\varepsilon/K$. If $K=0$, the <function> is constant. As usual for a <derivative> on this closed half-line, <continuity> at its endpoint is included.
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