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Mathematics LibreTexts

7.2: Proof of the Intermediate Value Theorem

  • Page ID
    7959
  • [ "article:topic", "Intermediate Value Theorem", "authorname:eboman" ]

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    Skills to Develop

    • Proof of Intermediate value theorem

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

    Theorem \(\PageIndex{1}\): Intermediate Value Theorem

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

    Sketch of Proof

    We have two cases to consider: \(f(a) ≤ v ≤ f(b)\) and \(f(a) ≥ v ≥ f(b)\).

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

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

    Exercise \(\PageIndex{1}\)

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

    Exercise \(\PageIndex{2}\)

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

    Exercise \(\PageIndex{3}\)

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

    Contributor

    • Eugene Boman (Pennsylvania State University) and Robert Rogers (SUNY Fredonia)