Proof of the Intermediate Value Theorem

Section 9.2 Proof of the Intermediate Value Theorem

We now have all of the tools we need to prove the Intermediate Value Theorem (IVT).

Theorem 9.2.1.

Intermediate Value Theorem (IVT)

Suppose \(f(x)\) is continuous on \([a,b]\) and \(v\) is any real number between \(f(a)\) and \(f(b)\text{.}\) Then there exists a real number \(c\in[\,a,b]\) such that \(f(c)=v\text{.}\)

Sketch of Proof.

We have two cases to consider: \(f(a)\leq v\leq f(b)\) and \(f(a)\geq v\geq f(b)\text{.}\)

We will look at the case \(f(a)\leq v\leq f(b)\text{.}\) Let \(x_1=a\) and \(y_1=b\text{,}\) so we have \(x_1\leq y_1\) and \(f(x_1)\leq v\leq f(y_1)\text{.}\) Let \(\,m_1\) be the midpoint of \([\,x_1,y_1]\) and notice that we have either \(f(m_1)\leq v\) or \(f(m_1)\geq v\text{.}\) If \(f(m_1)\leq v\) , then we relabel \(x_2=m_1\) and \(y_2=y_1\text{.}\) If \(f(m_1)\geq v\) , then we relabel \(x_2=x_1\) and \(y_2=m_1\text{.}\) In either case, we end up with \(x_1\leq x_2\leq y_2\leq y_1\text{,}\) \(y_2-x_2=\frac{1}{2}\left(y_1-x_1\right)\text{,}\) \(f(x_1)\leq v\leq f(y_1)\text{,}\) and \(f(x_2)\leq v\leq f(y_2)\text{.}\)

Now play the same game with the interval \([\,x_2,y_2]\text{.}\) If we keep playing this game, we will generate two sequences \(\left(x_n\right)\) and \(\left(y_n\right)\text{,}\) satisfying all of the conditions of the nested interval property. These sequences will also satisfy the following extra property: \(\forall\) \(n,\,f(x_n)\leq v\leq f(y_n)\text{.}\) By the NIP, there exists a \(c\) such that \(\,x_n\leq c\leq y_n\text{,}\) \(\forall\) \(n\text{.}\) This should be the \(c\) that we seek though this is not obvious. Specifically, we need to show that \(f(c)=v\text{.}\) This should be where the continuity of \(f\) at \(c\) and the extra property on \(\left(x_n\right)\)and \(\left(y_n\right)\) come into play.

Problem 9.2.2.

Turn the ideas of the previous paragraphs into a formal proof of the IVT for the case \(f(a)\leq v\leq f(b)\text{.}\)

Problem 9.2.3.

We can modify the proof of the case \(f(a)\leq v\leq f(b)\) into a proof of the IVT for the case \(f(a)\geq v\geq f(b)\text{.}\) However, there is a sneakier way to prove this case by applying the IVT to the function \(-f\text{.}\) Do this to prove the IVT for the case \(f(a)\geq v\geq f(b)\text{.}\)

Problem 9.2.4.

Use the IVT to prove that any polynomial of odd degree must have a real root.