= Solution
The <intermediate value theorem> states that, if $f$ is a <continuous function> on a <closed interval> $[a,b]$, every value between $f(a)$ and $f(b)$ equals $f(c)$ for some $c\in[a,b]$. If the value is strictly between the endpoint values, $c$ may be chosen in $(a,b)$.
Here is a proof from the <supremum> property of the real numbers. Suppose $f(a)<y<f(b)$; the reverse inequality is reduced to this case by replacing $f,y$ with $-f,-y$. Let
$$
E=\{t\in[a,b]:f(t)\leq y\},\qquad c=\sup E.
$$
The set is nonempty because it contains $a$ and is bounded above by $b$. <Continuity> at $a$ supplies points of $E$ to the right of $a$, while <continuity> at $b$ excludes a neighborhood of $b$ from $E$. Hence $a<c<b$. By the definition of <supremum>, there are $t_n\in E$ with $t_n\to c$. <Continuity> gives $f(c)=\lim f(t_n)\leq y$. If $f(c)<y$, <continuity> would give a point just to the right of $c$ still satisfying $f(t)<y$, contradicting that $c$ is an upper bound for $E$. Thus $f(c)=y$. Values equal to either endpoint are attained at that endpoint, completing the proof.
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