Intermediate Value Theorem

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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 $f \left({c}\right) = k$.


Corollary

Let $0 \in \R$ lie between $f \left({a}\right)$ and $f \left({b}\right)$.

That is, either:

$f \left({a}\right) < 0 < f \left({b}\right)$

or:

$f \left({b}\right) < 0 < f \left({a}\right)$


Then $f$ has a root in $\left({a \,.\,.\, b}\right)$.


Proof

This theorem is a restatement of Image of Interval by Continuous Function is Interval.


From Image of Interval by Continuous Function is Interval, 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.

$\blacksquare$


Also see


Sources