Intermediate Value Theorem

Theorem
Let $a,b\in\R$ such that $a<b$.

Let $\left[{a \,.\,.\, b}\right]$ be a closed interval.

Let $f: \left[{a \,.\,.\, b}\right] \to \R$ be a continuous real function.

Let $k \in \R$ lie between $f \left({a}\right)$ and $f \left({b}\right)$.

That is, either:
 * $f \left({a}\right) < k < f \left({b}\right)$

or:
 * $f \left({b}\right) < k < f \left({a}\right)$

Then $\exists c \in \left({a \,.\,.\, b}\right)$ such that $f \left({c}\right) = k$.

Proof
This theorem is a restatement of Continuous Image of Closed Interval is Closed Interval.

From Continuous Image of Closed Interval is Closed Interval, the image of $\left[{a \,.\,.\, b}\right]$ under $f$ is also a closed real interval.

Thus if $k$ lies between $f \left({a}\right)$ and $f \left({b}\right)$, it must be the case that:
 * $k \in \operatorname{Im} \left({\left({a \,.\,.\, b}\right)}\right)$

The result follows.

Also known as
This result is also known as Bolzano's theorem, for.

Some sources attribute it to, and call it the Weierstrass Intermediate Value Theorem.

Also see

 * Intermediate Value Theorem (Topology), of which this is a corollary