Question

The value \( c \) of Lagrange’s Mean Value Theorem for \( f(x) = e^x + 24 \) in \( [0,1] \) is:

• A. \( \log(e - 1) \)
• B. \( \log(e + 1) \)
• C. \( \log(e + 24) \)
• D. \( \log(e - 24) \)

Hint:

For functions of the form \( f(x) = e^x + C \), the LMVT condition simplifies to \( e^c = \frac{f(b) - f(a)}{b - a} \). Always check continuity and differentiability before applying LMVT.

Answer 1

The Correct Option is A

Step 1: Understanding Lagrange's Mean Value Theorem (LMVT) Lagrange’s 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 some \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a}. \] Here, we have the function: \[ f(x) = e^x + 24. \]
Step 2: Checking Continuity and Differentiability - \( f(x) = e^x + 24 \) is continuous for all real \( x \) because the exponential function is continuous. - \( f(x) \) is also differentiable for all real \( x \) because the derivative of \( e^x \) exists everywhere. Since both conditions are satisfied, LMVT is applicable in the given interval \([0,1]\).
Step 3: Finding \( f(a) \) and \( f(b) \) Let \( a = 0 \) and \( b = 1 \): \[ f(0) = e^0 + 24 = 1 + 24 = 25. \] \[ f(1) = e^1 + 24 = e + 24. \] The difference quotient is: \[ \frac{f(1) - f(0)}{1 - 0} = \frac{(e + 24) - 25}{1} = e - 1. \]
Step 4: Finding \( c \) Using LMVT The derivative of \( f(x) \) is: \[ f'(x) = \frac{d}{dx} (e^x + 24) = e^x. \] By LMVT, there exists some \( c \in (0,1) \) such that: \[ f'(c) = e^c = e - 1. \] Taking the natural logarithm on both sides: \[ c = \log(e - 1). \] Thus, the correct answer is: \[ \log(e - 1). \]