Intermediate Value Theorem
In short: Intermediate value theorem (IVT) guarantees that a connected, unbroken curve over [a, b] reaches each height lying between its starting and ending heights at least once on the way.
AP Calculus ABCalculusAP Calculus AB Unit 1
What is Intermediate Value Theorem?
The intermediate value theorem is a promise about unbroken graphs. If you can draw a curve from one end of an interval to the other without lifting your pencil, it has to cross every height between its starting height and its ending height.
Take \(f\) unbroken from \(x = a\) to \(x = b\). For every target \(N\) sitting between the outputs \(f(a)\) and \(f(b)\), the equation \(f(c) = N\) has a solution \(c\) somewhere in the open interval \((a, b)\). This intermediate value theorem proves existence only: it never counts the solutions or locates them.
How to spot it
- A problem asks you to show that an equation has a solution without actually solving it.
- The wording is “justify that there is a value c such that f(c) equals …” for a continuous function.
- A table lists function values at two inputs, and the target output sits strictly between them.
- You must argue that a polynomial has a zero inside a short interval such as [1, 2].
The method
- State that \(f\) is continuous on the closed interval \([a, b]\), and name the reason (polynomial, sum of continuous pieces, and so on).
- Compute \(f(a)\) and \(f(b)\) exactly.
- Check that the target value \(N\) lies strictly between \(f(a)\) and \(f(b)\).
- Conclude by naming the intermediate value theorem: some \(c\) in \((a, b)\) satisfies \(f(c) = N\).
Intermediate Value Theorem: solved example
Problem
Show that the equation \(x^3 + 2x = 11\) has a solution between 1 and 2.
Solution
Let \(f(x) = x^3 + 2x – 11\). Polynomials have no breaks, hence \(f\) is continuous across \([1, 2]\). Then \(f(1) = 1 + 2 – 11 = -8\) and \(f(2) = 8 + 4 – 11 = 1\).
Since \(-8 < 0 < 1\), zero lies between the two outputs. By the theorem, there is a number \(c\) in \((1, 2)\) with \(f(c) = 0\), which means \(c^3 + 2c = 11\) has a solution in \((1, 2)\).
Try it
Question 1
A continuous function \(g\) has \(g(0) = -3\) and \(g(6) = 10\). Must \(g(x) = 7\) for some \(x\) in \([0, 6]\)?
Show the answer
Yes. The number 7 lies between \(-3\) and 10, and \(g\) is continuous, so some \(x\) in \((0, 6)\) gives \(g(x) = 7\).
Question 2
Let \(h(x) = \frac{1}{x}\). Since \(h(-1) = -1\) and \(h(1) = 1\), can you conclude that \(h(x) = 0\) for some \(x\) in \([-1, 1]\)?
Show the answer
No. The function is not continuous at \(x = 0\), so the hypothesis fails. In fact \(h(x)\) is never 0.
Intermediate Value Theorem vs. Mean Value Theorem
The first theorem only needs continuity and promises that a chosen output is reached. The mean value theorem also needs differentiability, and it promises a point where the derivative equals the average rate of change \(\frac{f(b) – f(a)}{b – a}\). One speaks about heights, the other about slopes.
Keep going
Lessons
Worksheets and quizzes
- Continuity and the Intermediate Value Theorem – Worksheet
- Continuity and the Intermediate Value Theorem – Quiz
Related words
Whole dictionary: A–Z · AP Calculus AB · Calculus