Key takeaway: The intermediate value theorem says that a function continuous on a closed interval takes every value between its two endpoint values at least once somewhere inside that interval.
Grade 12 – AP Calculus ABAP Calculus AB Unit 1Calculus
Definition of Intermediate Value Theorem
The intermediate value theorem (IVT) describes what a continuous function cannot avoid doing. If a function has no breaks, jumps or holes on a closed interval, its graph is one unbroken piece, so it cannot travel from one height to another without passing through every height in between.
That makes the theorem a guarantee that a solution exists. It never tells you where the solution is or how many there are, only that at least one is there.
Formula and notation
Suppose \(f\) is continuous on \([a,b]\) and \(N\) is any number strictly between \(f(a)\) and \(f(b)\). Then
\[ \text{there is a number } c \text{ in } (a,b) \text{ with } f(c)=N. \]
Continuity on the whole interval is the key hypothesis. The target \(N\) may lie between the two endpoint outputs in either order.
Key facts
- Continuity on the closed interval is required; a jump can skip right over the target value.
- To prove an equation has a root, show that f changes sign between the endpoints.
- The theorem proves existence only; it does not find c or count how many c values work.
- Rising or falling graphs both qualify, as long as N sits between f(a) and f(b).
Intermediate Value Theorem example
Show that the equation \(x^3+2x-9=0\) has a solution between \(x=1\) and \(x=2\).
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Let \(f(x)=x^3+2x-9\). It is a polynomial, so it is continuous on \([1,2]\).
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Evaluate the endpoints: \(f(1)=1+2-9=-6\) and \(f(2)=8+4-9=3\).
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Since \(-6 < 0 < 3\), the value \(N=0\) lies between \(f(1)\) and \(f(2)\).
By the intermediate value theorem there is at least one number \(c\) in \((1,2)\) with \(f(c)=0\), so the equation has a solution in (1, 2).
Common mistake
Watch out
A student tests \(g(x)=\frac{1}{x-3}\) on \([2,4]\), finds \(g(2)=-1\) and \(g(4)=1\), and announces that \(g(x)=0\) must have a solution in \((2,4)\).
Why it is wrong
The sign change is real, but \(g\) has a vertical asymptote at \(x=3\), inside the interval, so it is not continuous there and the hypothesis of the theorem fails. In fact \(\frac{1}{x-3}\) is never zero. Always confirm continuity before using the intermediate value theorem.
Related terms and lessons
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