The intermediate value theorem (for short, the IVT) states that the range of a continuous function over a closed interval cannot have missing values. In other words, a continuous function takes any value between the values of the function at the endpoints of a closed interval inside the domain of the function.

#### A statement and a proof of the intermediate value theorem

**Result**. If \(f: [a,b] \rightarrow \mathbb{R}\) is a continuous and \(k\) is between \(f(a)\) and \(f(b)\), then there is a \(c \in (a,b)\) such that \(f(c) = k\).

The proof is as follows:

Without loss of generality, we may assume that $$f(a) < k < f(b).$$ The set \(S = \{x \in [a,b] : f(x) < k\}\) is nonempty since \(a\in S\), and also, bounded above by \(b\). So, the least upper bound of \(S\) exists. Let $$c = \sup(S).$$ It is clear that \(c \in [a,b)\).

Our claim is that \(f(c) = k\). On the contrary, assume that \(f(c) < k\) and set \(\epsilon = k – f(c)\). Therefore, there is a \(\delta > 0\) such that for all \(x\) with $$x \in [a,b] \text{ and } c < x < c+\delta,$$ we have $$\vert f(x) – f(c) \vert < k – f(c).$$ This implies that \(f(x) < k\) which means that \(x \in S\) and this is a contradiction because \(c\) is the least upper bound of \(S\).

A similar argument shows that \(f(c) > k\) is also impossible. Hence, \(f(c) = k\), as required.

#### An essential corollary to the intermediate value theorem

An essential corollary to the intermediate value theorem is the Bolzano’s theorem which states that if a function is continuous on a closed interval and is sometimes positive and sometimes negative, then it must be zero at some point. More precisely:

**Result** (Bolzano’s theorem). Let \(f: [a,b] \rightarrow \mathbb{R}\) be continuous with \(f(a)f(b) < 0\). Then, there is a \(c \in (a,b)\) such that \(f(c) = 0\).

**Exercise**. Prove that every polynomial of odd degree has at least one real root.

**Solution**. Let \(p(x) = \sum_{i=0}^{n} a_i x^i\) be a polynomial of degree \(n\), where \(n\) is an odd number and \(a_n\) a nonzero real number. Without loss of generality, we may assume that \(a_n\) is positive. It is clear that we may choose \(b\) large enough such that \(p(b) > 0\). Also, we may choose \(a\) small enough such that \(p(a) < 0\). Now, by the Bolzano’s theorem, there is a \(c\) such that \(p(c) = 0\) and the proof is complete.