AP Calculus BC 5.1 Using the Mean Value Theorem Study Notes - New Syllabus
AP Calculus BC 5.1 Using the Mean Value Theorem Study Notes- New syllabus
AP Calculus BC 5.1 Using the Mean Value Theorem Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- A function’s derivative can be used to understand some behaviors of the function.
Key Concepts:
- Mean Value Theorem (MVT)
Mean Value Theorem (MVT)
Mean Value Theorem (MVT)
Theorem Statement:
If a function \( f(x) \) is:
- Continuous on the closed interval \([a, b]\), and
- Differentiable on the open interval \((a, b)\),
then there exists at least one point \( c \in (a, b) \) such that:
\( f'(c) = \dfrac{f(b) – f(a)}{b – a} \)
This means the instantaneous rate of change (the derivative at some point \( c \)) equals the average rate of change over the interval \([a, b]\).
Geometric Interpretation:
There is at least one point on the curve \( f(x) \) between \( a \) and \( b \) where the tangent line is parallel to the secant line connecting \( (a, f(a)) \) and \( (b, f(b)) \).
Average rate of change: \( \dfrac{f(b) – f(a)}{b – a} \) — slope of the secant line
Instantaneous rate of change: \( f'(c) \) — slope of the tangent line at some point \( c \)
Example:
Let \( f(x) = x^2 \) on the interval \([1, 3]\). Show that the Mean Value Theorem applies and find the value(s) of \( c \) that satisfy the theorem.
▶️Answer/Explanation
\( f(x) = x^2 \) is continuous and differentiable everywhere, so MVT applies.
Average rate of change:
\( \frac{f(3) – f(1)}{3 – 1} = \frac{9 – 1}{2} = 4 \)
Find \( c \) such that \( f'(c) = 4 \):
\( f'(x) = 2x \Rightarrow 2c = 4 \Rightarrow c = 2 \)
So, \( c = 2 \) satisfies the MVT.
Example:
Let \( f(x) = \ln(x) \) on the interval \([1, e] \). Show that the Mean Value Theorem applies and find a value \( c \in (1, e) \) such that \( f'(c) = \dfrac{f(e) – f(1)}{e – 1} \).
▶️Answer/Explanation
\( f(x) = \ln x \) is continuous and differentiable for \( x > 0 \), so MVT applies.
Average rate of change:
\( \frac{\ln(e) – \ln(1)}{e – 1} = \frac{1 – 0}{e – 1} \)
We want \( f'(c) = \dfrac{1}{e – 1} \). Since \( f'(x) = \dfrac{1}{x} \), we solve:
\( \frac{1}{c} = \frac{1}{e – 1} \Rightarrow c = e – 1 \)
So, \( c = e – 1 \in (1, e) \) satisfies the theorem.
Example:
Let \( f(x) = \dfrac{x^2}{2} – x \) be defined on the interval \([0, 4]\).
Which of the following values of \( c \) in the open interval \( (0, 4) \) satisfies the conclusion of the Mean Value Theorem?
- \( c = 1 \)
- \( c = 2 \)
- \( c = 3 \)
- \( c = 4 \)
▶️Answer/Explanation
The function \( f(x) = \dfrac{x^2}{2} – x \) is a polynomial, so it’s continuous and differentiable everywhere. MVT applies on \([0, 4]\).
Compute the average rate of change:
\( \frac{f(4) – f(0)}{4 – 0} = \frac{\left(\frac{16}{2} – 4\right) – (0)}{4} = \frac{8 – 4}{4} = 1 \)
Step 2: Find \( c \in (0, 4) \) such that \( f'(c) = 1 \)
\( f'(x) = x – 1, \quad \text{so set } f'(c) = 1 \Rightarrow c – 1 = 1 \Rightarrow c = 2 \)
Correct answer: B. \( c = 2 \)
Example:
Let \( f(x) = 2 \sin x \). The graph of \( f \) crosses the origin at point \( A(0, 0) \) and point \( B\left( \frac{\pi}{2}, 2 \right) \).
Find the \( x \)-coordinate of the point on the graph of \( f \), between points \( A \) and \( B \), at which the line tangent to the graph is parallel to the line \( AB \). Round or truncate to three decimals.
▶️Answer/Explanation
We apply the Mean Value Theorem (MVT) since the function \( f(x) = 2\sin x \) is continuous and differentiable on the interval \( [0, \frac{\pi}{2}] \).
Step 1: Compute the average rate of change from \( A \) to \( B \):
\( \frac{f\left( \frac{\pi}{2} \right) – f(0)}{\frac{\pi}{2} – 0} = \frac{2 – 0}{\frac{\pi}{2}} = \frac{2}{\frac{\pi}{2}} = \frac{4}{\pi} \)
Step 2: Find where the derivative equals this value. Since \( f(x) = 2\sin x \), then:
\( f'(x) = 2\cos x \)
Set \( f'(x) = \frac{4}{\pi} \):
\( 2\cos x = \frac{4}{\pi} \quad \Rightarrow \quad \cos x = \frac{2}{\pi} \)
Step 3: Solve for \( x \):
\( x = \cos^{-1} \left( \frac{2}{\pi} \right) \approx \cos^{-1}(0.6366) \approx 0.884 \)
Final Answer: \( \boxed{0.884} \)
Example:
Let \( g \) be a continuous function. The graph of the piecewise-linear function \( g’ \), the derivative of \( g \), is shown for \( -4 \leq x \leq 4 \).
Find the average rate of change of \( g'(x) \) on the interval \( -4 \leq x \leq 4 \).
Does the Mean Value Theorem applied on the interval \( -4 \leq x \leq 4 \) guarantee a value of \( c \), for \( -4 < x < 4 \), such that \( g”(c) \) is equal to this average rate of change? Why or why not?
▶️Answer/Explanation
Average rate of change of \( g'(x) \):
From the graph:
- At \( x = -4 \), \( g'(-4) = 2 \)
- At \( x = 4 \), \( g'(4) = -2 \)
Average rate of change of \( g'(x) \) from \( x = -4 \) to \( x = 4 \) is:
\( \frac{g'(4) – g'(-4)}{4 – (-4)} = \frac{-2 – 2}{8} = \frac{-4}{8} = -0.5 \)
- The MVT states: if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a \( c \in (a, b) \) such that:
\( g”(c) = \frac{g'(b) – g'(a)}{b – a} \)
- But here, we are dealing with \( g'(x) \), the derivative of \( g \). So, applying MVT to \( g’ \) requires \( g’ \) to be continuous and differentiable on the interval \((-4, 4)\).
- From the graph, \( g'(x) \) is piecewise-linear and not differentiable at \( x = -1 \), \( x = 2 \), and possibly at corners. Thus, \( g’ \) is not differentiable on the entire open interval.
- Therefore, since \( g’ \) is not differentiable on \((-4, 4) \), the Mean Value Theorem does not apply to \( g’ \), and we cannot guarantee the existence of \( c \) such that \( g”(c) = -0.5 \).
Final Answer: The average rate of change is \( -0.5 \), but the MVT does not apply to \( g’ \) on this interval because \( g’ \) is not differentiable throughout \( (-4, 4) \). So no such \( c \) is guaranteed to exist.