AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/1/1-16)
Unit 1 · Topic 1.16
1.16 Working with the Intermediate Value Theorem (IVT)
The Intermediate Value Theorem (IVT) says a continuous function on a closed interval can't skip any value between its endpoint outputs. It guarantees that a value is reached somewhere, without telling you where. You must confirm continuity before you use it.
Key terms
- Intermediate Value Theorem
- continuous on a closed interval
- existence theorem
- hypotheses (conditions)
- zero of a function
The theorem, with its conditions
If f is continuous on the closed interval [a, b], and N is any number between f(a) and f(b), then there is at least one number c in (a, b) with f(c) = N.
In picture form: a continuous curve that starts at height f(a) and ends at height f(b) must cross every horizontal line in between. It can't jump over one.
A common use is finding zeros: if f is continuous on [a, b] and f(a) and f(b) have opposite signs, then 0 is between them, so f(c) = 0 for some c in (a, b).
What it does and doesn't tell you
Read the theorem carefully. It is a promise about existence only:
- It guarantees existence: at least one c works. There might be many.
- It doesn't tell you where c is or how many there are.
- It needs continuity on the whole closed interval. Without continuity, the conclusion can fail.
- It only promises values between f(a) and f(b). The function can also go outside that range; the theorem doesn't care.
The justification that earns the point
A complete IVT justification has three parts. First, state that f is continuous on [a, b], with a reason if one is available (for example, it's a polynomial, or it's differentiable). Second, show the value N is between f(a) and f(b) with actual numbers. Third, conclude with the theorem's name: “By the Intermediate Value Theorem, there is a value c in (a, b) such that f(c) = N.”
With tables, IVT can show a minimum number of solutions. Each time the table's values cross N between two consecutive inputs, you get at least one solution in that subinterval.
IVT in context
IVT often appears with a real quantity measured at a few times. Say a room's temperature T(t), in °F, is continuous, with T(0) = 64 and T(3) = 75. Then IVT guarantees that at some time between t = 0 and t = 3 the temperature was exactly 70°F, because 70 is between 64 and 75.
The converse isn't true. If f(c) = N for some c in the interval, that doesn't mean f is continuous. A function with a jump can still happen to hit N.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Showing a root exists
Show that f(x) = x³ + x − 3 has a zero between x = 1 and x = 2.
Show the solutionHide the solution
- Step 1: f is a polynomial, so it is continuous on [1, 2].
- Step 2: f(1) = 1 + 1 − 3 = −1 and f(2) = 8 + 2 − 3 = 7.
- Step 3: Since −1 < 0 < 7, the value 0 is between f(1) and f(2).
Answer: By the Intermediate Value Theorem, there is a value c in (1, 2) with f(c) = 0. (A calculator shows c ≈ 1.213, but IVT alone only guarantees it exists.)
- Example 2
Minimum number of solutions from a table
f is continuous on [0, 5]. f(0) = 2, f(2) = −1 and f(5) = 4. What is the least number of solutions f(x) = 0 must have on [0, 5]?
Show the solutionHide the solution
- Step 1: On [0, 2]: f goes from 2 to −1, and 0 is between them. f is continuous there, so IVT gives at least one zero in (0, 2).
- Step 2: On [2, 5]: f goes from −1 to 4, so IVT gives at least one zero in (2, 5).
- Step 3: These intervals don't overlap, so the zeros are different.
Answer: At least 2 solutions.
- Example 3
Trap: no continuity, no guarantee
For g(x) = 1/x, g(−1) = −1 and g(1) = 1. A student says IVT guarantees g(c) = 0 for some c in (−1, 1). Is that right?
Show the solutionHide the solution
- Step 1: Check the condition: g is not continuous on [−1, 1], because it is undefined at x = 0 (a vertical asymptote).
- Step 2: So IVT does not apply.
- Step 3: In fact 1/x is never 0, so there is no such c.
Answer: No. IVT requires continuity on [−1, 1], which fails at x = 0, and 1/x = 0 has no solution.
Common mistakes
- Skipping the continuity statement. On free response, you lose the point if you don't say f is continuous on the closed interval.
- Saying “f(c) = 0 at c = 1.5” or naming a c. IVT only says some c exists.
- Mixing up IVT with the Mean Value Theorem. IVT is about function values; MVT (5.1) is about slopes.
On the exam
- IVT shows up in free response with a table of values: “Must there be a value c where f(c) = 5? Justify.” Name the interval, state continuity, show the numbers bracket 5 and name the theorem.
- If the problem says f is differentiable, you can say “f is differentiable, so it is continuous” to meet the condition.
Connected topics
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Check yourself
4 questions on 1.16 Working with the Intermediate Value Theorem (IVT). Pick an answer to see if you got it, and why.
| x | f(x) |
|---|---|
| 0 | 3 |
| 1 | −1 |
| 2 | 2 |
| 3 | 5 |
| 4 | −2 |
Selected values of a continuous function f
The function f is continuous on the closed interval [0, 4], and selected values of f are given in the table. What is the least number of zeros that f must have on [0, 4] ?
For the function f described above, on which of the following intervals must there be a value c with f(c) = −3/2 ?
Let f(x) = 1/(x − 2). Then f(1) = −1 and f(3) = 1, but f(x) ≠ 0 for every x in [1, 3]. Why does this not contradict the Intermediate Value Theorem?
Let f(x) = x³ + x − 5. Because f is continuous with f(1) = −3 and f(2) = 5, the Intermediate Value Theorem guarantees that f(c) = 0 for some c in the interval (1, 2). What is the value of c ?
0 of 4 answered