Intermediate Value Theorem

Theorem
Let $I$ be a real interval.

Let $a, b \in I$ such that $\left({a \,.\,.\, b}\right)$ is an open interval.

Let $f: I \to \R$ be a real function which is continuous on $\left({a \,.\,.\, b}\right)$.

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 Image of Interval by Continuous Function.

From Image of Interval by Continuous Function, the image of $\left({a \,.\,.\, b}\right)$ under $f$ is also a real interval (but not necessarily open).

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 see

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