AP Calculus BC 1.16 Working with the Intermediate Value Theorem (IVT) Study Notes - New Syllabus
AP Calculus BC 1.16 Working with the Intermediate Value Theorem (IVT) Study Notes- New syllabus
AP Calculus BC 1.16 Working with the Intermediate Value Theorem (IVT) Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Explain the behavior of a function on an interval using the Intermediate Value Theorem.
Key Concepts:
- Intermediate Value Theorem (IVT)
Intermediate Value Theorem (IVT)
Intermediate Value Theorem (IVT)
The Intermediate Value Theorem states:
If \( f(x) \) is continuous on the closed interval \( [a, b] \), and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one number \( c \in (a, b) \) such that \( f(c) = N \).
Key Points:
- The function must be continuous on the entire interval \( [a, b] \).
- The theorem guarantees existence of a root (or any value \( N \)) but does not tell us how to find it exactly.
- Often used to prove the existence of zeros of a function.
Graphical Idea: If a continuous function moves from \( f(a) \) to \( f(b) \), it must pass through every intermediate value between them.
Applications:
- Root existence: If \( f(a) \) and \( f(b) \) have opposite signs, there is at least one root in \( (a, b) \).
- Ensuring a function takes on a particular value within an interval.
Example:
Show that the equation \( x^3 + x – 4 = 0 \) has a root in the interval \( [1, 2] \).
▶️ Answer/Explanation
Define \( f(x) = x^3 + x – 4 \). The function is a polynomial, so it is continuous on \( [1, 2] \).
Compute \( f(1) \) and \( f(2) \):
\( f(1) = 1^3 + 1 – 4 = -2 \), \( f(2) = 2^3 + 2 – 4 = 6 \).
\( f(1) < 0 \), \( f(2) > 0 \). Since \( f(x) \) is continuous, by IVT there exists \( c \in (1, 2) \) such that \( f(c) = 0 \).
Answer: A root exists in \( (1, 2) \).
Example:
The temperature of a liquid cools from \( 80^\circ C \) at 2 PM to \( 60^\circ C \) at 3 PM. Use IVT to show there was a time when the temperature was \( 70^\circ C \).
▶️ Answer/Explanation
Let \( T(t) \) be the temperature at time \( t \), which is continuous.
At 2 PM, \( T(2) = 80^\circ C \); at 3 PM, \( T(3) = 60^\circ C \).
\( 70^\circ C \) lies between \( 60^\circ C \) and \( 80^\circ C \). By IVT, there exists a time \( c \in (2, 3) \) where \( T(c) = 70^\circ C \).
Answer: Yes, there was a time between 2 PM and 3 PM when the temperature was \( 70^\circ C \).
Example:
Verify if the equation \( \cos x = x \) has a solution in the interval \( [0, 1] \).
▶️ Answer/Explanation
Define \( f(x) = \cos x – x \). The function is continuous on \( [0, 1] \).
Compute \( f(0) = \cos 0 – 0 = 1 \), \( f(1) = \cos 1 – 1 \approx -0.46 \).
\( f(0) > 0 \), \( f(1) < 0 \). By IVT, there exists \( c \in (0, 1) \) such that \( f(c) = 0 \), i.e., \( \cos c = c \).
Answer: Yes, a solution exists in \( (0, 1) \).
Example:
Show that the equation \( x^5 – x^2 + 1 = 0 \) has a root in the interval \( [-2, -1] \).
▶️ Answer/Explanation
Define \( f(x) = x^5 – x^2 + 1 \). This is a polynomial, so it is continuous on \( [-2, -1] \).
Compute \( f(-2) \) and \( f(-1) \):
\( f(-2) = (-2)^5 – (-2)^2 + 1 = -32 – 4 + 1 = -35 \).
\( f(-1) = (-1)^5 – (-1)^2 + 1 = -1 – 1 + 1 = -1 \).
Notice both values are negative (\( f(-2) = -35, f(-1) = -1 \)), so no sign change. Try a point between them, say \( x = 0 \):
\( f(0) = 0^5 – 0^2 + 1 = 1 \).
So \( f(-1) = -1 \), \( f(0) = 1 \) → sign change occurs on \( (-1, 0) \).
Conclusion: By IVT, there exists \( c \in (-1, 0) \) such that \( f(c) = 0 \).
Answer: A root exists in \( (-1, 0) \).
Example:
Let \( f \) be a continuous function such that \( f(1) = 7 \) and \( f(7) = 1 \). Let \( g \) be the function given by \( g(x) = f(x) – x \).
Explain why there must be a value \( c \) for \( 1 < c < 7 \) such that \( g(c) = 0 \).
▶️ Answer/Explanation
Since \( f(x) \) is continuous, and \( g(x) = f(x) – x \) is the difference of two continuous functions, \( g(x) \) is also continuous on \( [1, 7] \).
Compute \( g(1) \) and \( g(7) \):
\( g(1) = f(1) – 1 = 7 – 1 = 6 \) (positive)
\( g(7) = f(7) – 7 = 1 – 7 = -6 \) (negative)
Since \( g(x) \) changes sign from positive to negative and is continuous on \( [1, 7] \), by the Intermediate Value Theorem there exists \( c \in (1, 7) \) such that \( g(c) = 0 \).
Answer: Yes, there is a value \( c \) in \( (1, 7) \) where \( g(c) = 0 \).