Short on time? This page is the recap. For explanations, the graphs and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- Theorem. If f is continuous on the closed interval [a, b] and d is between f(a) and f(b), then there is at least one c between a and b with f(c) = d.
- It is an existence theorem: it does not say where.
- “At least one”: there may be several values of c.
- If a condition fails, the theorem makes no promise. The value may still occur, or it may be skipped.
- To show g(x) = h(x) has a solution, apply the theorem to g(x) − h(x) with target 0.
Key relationships
| Situation | What you can conclude |
|---|---|
| f continuous on [a, b], f(a) and f(b) of opposite signs | At least one zero in (a, b) |
| f continuous on [a, b], d strictly between f(a) and f(b) | At least one c in (a, b) with f(c) = d |
| f continuous, table values trap d in k separate sub-intervals | At least k solutions of f(x) = d |
| f(a) and f(b) on the same side of d | Nothing: the theorem is silent |
| f not continuous somewhere in [a, b] | Nothing: the theorem may not be used |
Continuity reasons you can quote: polynomial; sum, product or composition of continuous functions; quotient where the denominator is never 0 on [a, b]; “given in the question”; differentiable on the interval (Unit 2). For a piecewise function, check every join.
The three-part justification
- f is continuous on [a, b] because …
- f(a) = …, f(b) = …, and d is between them (write the inequality).
- Therefore, by the Intermediate Value Theorem, there is a c in (a, b) with f(c) = d.
Assumptions behind the method
- Continuity must hold on the whole closed interval, not only at the endpoints.
- The target must be between the end values f(a) and f(b), not merely between a and b.
- Different solutions counted from a table must come from sub-intervals that do not overlap.
Mistakes to avoid
- No continuity statement, or “continuous” with no reason.
- No numbers: “by IVT, yes” without f(a), f(b) and the inequality.
- Claiming exactly one c, or claiming to know where c is.
- Saying “no solution” when a condition fails. The right phrase is “not guaranteed”.
- Using the theorem across a jump in a piecewise function.
- Treating a table as a full count: it gives a minimum number only.
Quick self-check
- q(x) = x³ + 2x − 5. Show q has a zero in (1, 2). (q is a polynomial, so continuous on [1, 2]. q(1) = −2 < 0 < 7 = q(2). By the Intermediate Value Theorem, q(c) = 0 for some c in (1, 2).)
- g is continuous on [0, 4] with g(0) = 7 and g(4) = 2. Is g(c) = 3 guaranteed for some c in (0, 4)? Is g(c) = 8? (Yes for 3, because 2 < 3 < 7. Not guaranteed for 8, because 8 is not between 2 and 7.)
- f(x) = 1/x, f(−1) = −1, f(1) = 1. But 1/x is never 0. Does this contradict the theorem? (No: f is not continuous at x = 0, so the theorem does not apply to [−1, 1].)
Next: practice questions.