Problem 9: Let \( f(x)=\ln (x) \) on the interval \( [1, e] \). . Find the value of \( c \) in \( (1, e) \) where \( f^{\prime}(c)=\frac{f(e)-f(1)}{e-1} \)

To solve for the value of \( c \) in the interval \( (1, e) \) that satisfies the Mean Value Theorem (MVT), follow these steps: **Given:** - Function: \( f(x) = \ln(x) \) - Interval: \( [1, e] \) **Mean Value Theorem Statement:** There exists at least one \( c \) in \( (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] **Applying MVT to the given function:** 1. **Compute \( f(e) \) and \( f(1) \):** \[ f(e) = \ln(e) = 1 \quad \text{and} \quad f(1) = \ln(1) = 0 \] 2. **Calculate the difference quotient:** \[ \frac{f(e) - f(1)}{e - 1} = \frac{1 - 0}{e - 1} = \frac{1}{e - 1} \] 3. **Find the derivative \( f'(x) \):** \[ f'(x) = \frac{d}{dx} \ln(x) = \frac{1}{x} \] 4. **Set the derivative equal to the difference quotient:** \[ f'(c) = \frac{1}{c} = \frac{1}{e - 1} \] 5. **Solve for \( c \):** \[ \frac{1}{c} = \frac{1}{e - 1} \quad \Rightarrow \quad c = e - 1 \] **Verification:** Since \( e \approx 2.718 \), then \( e - 1 \approx 1.718 \), which lies within the interval \( (1, e) \). **Conclusion:** The value of \( c \) that satisfies the Mean Value Theorem for the function \( f(x) = \ln(x) \) on the interval \( [1, e] \) is: \[ c = e - 1 \] **Answer:** After simplifying, the required value is e minus one. Thus, c = e − 1.