AP Calculus AB 5.1 The Mean Value Theorem if f(x) = l(x) on [1, e^2]

### AP Calculus AB 5.1: The Mean Value Theorem for \( f(x) = \ln(x) \) on \([1, e^2]\)

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### **1. Statement of the Mean Value Theorem (MVT)**:
The **Mean Value Theorem** states that 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:
\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]
This means the instantaneous rate of change (slope of the tangent line) at some point \( c \) is equal to the average rate of change of the function over the interval \([a, b]\).

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### **2. Verifying the Conditions for \( f(x) = \ln(x) \)**:
- \( f(x) = \ln(x) \) is **continuous** on \([1, e^2]\) because the natural logarithm function is continuous for all \( x is greater than 0 \).
- \( f(x) = \ln(x) \) is also **differentiable** on the open interval \( (1, e^2) \) because the derivative \( f'(x) = \frac{1}{x} \) exists for all \( x is greater than 0 \).

Thus, the function satisfies the conditions of the Mean Value Theorem on the interval \([1, e^2]\).

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### **3. Applying the Mean Value Theorem**:

#### **Step 1: Calculate the Average Rate of Change**:
The average rate of change of \( f(x) = \ln(x) \) on the interval \([1, e^2]\) is:
\[
\frac{f(e^2) - f(1)}{e^2 - 1}
\]

1. **Evaluate \( f(e^2) \)**:
\[
f(e^2) = \ln(e^2) = 2
\]

2. **Evaluate \( f(1) \)**:
\[
f(1) = \ln(1) = 0
\]

3. **Find the average rate of change**:
\[
\frac{f(e^2) - f(1)}{e^2 - 1} = \frac{2 - 0}{e^2 - 1} = \frac{2}{e^2 - 1}
\]

#### **Step 2: Find the Derivative of \( f(x) \)**:
The derivative of \( f(x) = \ln(x) \) is:
\[
f'(x) = \frac{1}{x}
\]

#### **Step 3: Set the Derivative Equal to the Average Rate of Change**:
Using the Mean Value Theorem, set \( f'(c) = \frac{1}{c} \) equal to the average rate of change:
\[
\frac{1}{c} = \frac{2}{e^2 - 1}
\]

#### **Step 4: Solve for \( c \)**:
Solving for \( c \):
\[
c = \frac{e^2 - 1}{2}
\]

Thus, the value of \( c \) that satisfies the Mean Value Theorem is:
\[
c = \frac{e^2 - 1}{2}
\]

Since \( c \approx 3.1945 \), this value lies within the interval \( (1, e^2) \), which is approximately \( (1, 7.3891) \), satisfying the theorem's conditions.

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### **4. Conclusion**:
For \( f(x) = \ln(x) \) on the interval \([1, e^2]\), the Mean Value Theorem guarantees a point \( c = \frac{e^2 - 1}{2} \) in the interval \( (1, e^2) \) where the instantaneous rate of change (slope of the tangent line) equals the average rate of change over the interval. This point can be interpreted as the location where the behavior of the function aligns locally with its overall behavior across the interval.

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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