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Intermediate Value
Theorem

Master the IVT: understand when a continuous function is guaranteed to hit every value between two outputs.

AP Calculus AB

What Is the Intermediate Value Theorem?

The Intermediate Value Theorem says that if a function \( f \) is continuous on a closed interval \( [a, b] \), and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one value \( c \) in \( (a, b) \) such that \( f(c) = N \).

A continuous route through hills: elevation, grade, and intermediate height

A walking path crosses a ridge and hollow beside a staff-bearing observer.A continuous function takes every value between its endpoint values. Identify a target height between them before applying the Intermediate Value Theorem.

Intermediate Value Theorem: a height

A continuous function takes every value between its endpoint values. Identify a target height between them before applying the Intermediate Value Theorem.A continuous function takes every value between its endpoint values. Identify a target height between them before applying the Intermediate Value Theorem.

The Continuity Requirement

The IVT only applies when \( f \) is continuous on the entire closed interval \( [a, b] \). A single discontinuity—a jump, a hole, or an asymptote—can invalidate the conclusion.

✓ Continuous
Polynomials, \( \sin x \), \( e^x \), and compositions of continuous functions on \( [a,b] \).
✕ Not Continuous
Piecewise functions with jumps, \( \frac{1}{x} \) at \( x = 0 \), or functions with removable holes in \( [a,b] \).

Visual Intuition

x y \( N \) \( a \) \( b \) \( c \) \( f(a) \) \( f(b) \) \( f(c) = N \)

The Formal Statement

1
Hypothesis 1: \( f \) is continuous on the closed interval \( [a, b] \).
2
Hypothesis 2: \( N \) lies strictly between \( f(a) \) and \( f(b) \): that is, \( \min(f(a), f(b)) < N < \max(f(a), f(b)) \).
3
Conclusion: There exists at least one \( c \in (a, b) \) with \( f(c) = N \).

Existence, Not Location

A critical detail: the IVT guarantees that a value \( c \) exists, but it does not tell you what \( c \) is. You know the function hits \( N \) somewhere in \( (a, b) \), but the theorem alone cannot pinpoint where.

IVT Says
"There is at least one \( c \) where \( f(c) = N \)."
IVT Does Not Say
"The value is \( c = 2.5 \)" or "there is exactly one such \( c \)."

IVT vs EVT vs MVT

IVT EVT MVT
Requires Continuous on \( [a,b] \) Continuous on \( [a,b] \) Continuous on \( [a,b] \), differentiable on \( (a,b) \)
Guarantees \( f(c) = N \) for some \( c \) Absolute max and min exist \( f'(c) = \frac{f(b)-f(a)}{b-a} \)
About Output values Extreme values Slope / rate of change

How to Apply the IVT

1
Verify continuity. State that \( f \) is continuous on \( [a, b] \) and explain why (polynomial, composition of continuous functions, etc.).
2
Compute the endpoints. Find \( f(a) \) and \( f(b) \).
3
Identify the target value \( N \). Confirm that \( N \) lies between \( f(a) \) and \( f(b) \).
4
State the conclusion. "By the IVT, there exists at least one \( c \in (a, b) \) such that \( f(c) = N \)."

Example 1: Finding a Root

Show that \( f(x) = x^3 - 4x + 1 \) has a root on the interval \( [1, 2] \).

1
Continuity: \( f(x) = x^3 - 4x + 1 \) is a polynomial, so it is continuous everywhere, including on \( [1, 2] \).
2
Endpoints: \( f(1) = 1 - 4 + 1 = \) \( -2 \) and \( f(2) = 8 - 8 + 1 = \) \( 1 \).
3
Target: We want \( f(c) = 0 \). Since \( \textcolor{#f87171}{-2} < 0 < \textcolor{#4ade80}{1} \), the value \( N = 0 \) is between \( f(1) \) and \( f(2) \).
4
Conclusion: By the IVT, there exists at least one \( c \in (1, 2) \) such that \( f(c) = 0 \).

Example 1: Visualized

x y \( y=0 \) 1 2 \( f(1)=-2 \) \( f(2)=1 \) root

Example 2: A Specific Value

Let \( g(x) = e^x - 3 \). Show there exists a \( c \in (0, 2) \) such that \( g(c) = 2 \).

1
Continuity: \( g(x) = e^x - 3 \) is continuous on \( [0, 2] \) because \( e^x \) is continuous everywhere.
2
Endpoints: \( g(0) = e^0 - 3 = 1 - 3 = \) \( -2 \). \( g(2) = e^2 - 3 \approx 7.389 - 3 = \) \( 4.389 \).
3
Target: \( N = 2 \). Since \( \textcolor{#f87171}{-2} < 2 < \textcolor{#4ade80}{4.389} \), the value \( 2 \) is between \( g(0) \) and \( g(2) \).
4
Conclusion: By the IVT, there exists \( c \in (0, 2) \) such that \( g(c) = 2 \).

Example 3: Using a Table

The table below gives values of a continuous function \( h(x) \). Must there be a value \( c \) in \( [2, 5] \) with \( h(c) = 0 \)?

\( x \) 2 3 4 5
\( h(x) \) \( -3 \) \( -1 \) \( 4 \) \( 7 \)

When the IVT Does Not Apply

x y \( N \) \( a \) \( b \) jump! \( N \) is skipped

IVT and Sign Changes

The most common AP application: showing an equation \( f(x) = 0 \) has a solution. If \( f \) is continuous and \( f(a) \) and \( f(b) \) have opposite signs, the IVT guarantees a root in \( (a, b) \).

\( f(a) < 0 \)
Function is negative at the left endpoint.
\( f(b) > 0 \)
Function is positive at the right endpoint.
Root exists
Since \( 0 \) is between a negative and positive value, IVT gives a root.

Common Mistakes

✕ Forgetting Continuity
Applying IVT without stating the function is continuous on \( [a, b] \). This costs points on free-response questions.
✕ Claiming Uniqueness
Saying "there is exactly one \( c \)." The IVT guarantees at least one—there could be more.
✕ Open vs Closed
The function must be continuous on the closed interval \( [a,b] \). The guaranteed \( c \) lives in the open interval \( (a,b) \).

Quick Reference

Element Details
Hypothesis \( f \) continuous on \( [a, b] \); \( N \) between \( f(a) \) and \( f(b) \)
Conclusion There exists \( c \in (a, b) \) with \( f(c) = N \)
Guarantees Existence (at least one), not uniqueness or location
Most common use Proving a root exists via a sign change
AP tip Always state continuity and show \( N \) is between endpoints

Question 1 of 5

If \( f \) is continuous on \( [1, 5] \), \( f(1) = -3 \), and \( f(5) = 7 \), which value is the IVT guaranteed to be attained by \( f \) on \( (1, 5) \)?

Question 2 of 5

A student writes: "Since \( f(2) = -1 \) and \( f(6) = 5 \), by IVT there exists \( c \in (2, 6) \) with \( f(c) = 0 \)." What critical piece is missing from this argument?

Question 3 of 5

Let \( p(x) = x^4 - 6x^2 + 2 \). Which interval is guaranteed by the IVT to contain a root of \( p \)?

Question 4 of 5

The IVT tells us that if \( f \) is continuous on \( [a,b] \) and \( N \) is between \( f(a) \) and \( f(b) \), then \( f(c) = N \) for some \( c \). Which of the following is true about \( c \)?

Question 5 of 5

A function \( f \) is defined on \( [0, 4] \) with \( f(0) = 1 \) and \( f(4) = 1 \). A student claims: "By the IVT, \( f \) must equal \( 3 \) somewhere on \( (0, 4) \)." Is the student correct?