Given `f^(prime)(1)=1` and `d/(dx)(f(2x))=f^(prime)(x)AAx > 0`.If `f^(prime)(x)` is differentiable then there exits a number `c in (2,4)` such that `f''(c)` equals

Options

A. `1/4`
B. `(-1)/(2)`
C. `-1(1)/(4)`
D. `-1/8`

Answer

To solve the given problem step-by-step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the Given Information
We are given:
1. \( f'(1) = 1 \)
2. \( \frac{d}{dx} f(2x) = f'(x) \) for all \( x > 0 \)
3. \( f'(x) \) is differentiable.

### Step 2: Differentiate \( f(2x) \)
Using the chain rule, we differentiate \( f(2x) \):
\[ \frac{d}{dx} f(2x) = 2f'(2x) \]
According to the problem, this is equal to \( f'(x) \). Therefore, we have:
\[ 2f'(2x) = f'(x) \]

### Step 3: Analyze the Equation
From the equation \( 2f'(2x) = f'(x) \), we can express \( f'(2x) \) in terms of \( f'(x) \):
\[ f'(2x) = \frac{1}{2} f'(x) \]

### Step 4: Find Values of \( f'(2) \) and \( f'(4) \)
1. **For \( x = 1 \)**:
\[ f'(2) = \frac{1}{2} f'(1) = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

2. **For \( x = 2 \)**:
\[ f'(4) = \frac{1}{2} f'(2) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

### Step 5: Apply Lagrange's Mean Value Theorem
According to the Mean Value Theorem, there exists a number \( c \) in the interval \( (2, 4) \) such that:
\[ f''(c) = \frac{f'(4) - f'(2)}{4 - 2} \]

### Step 6: Substitute the Values
Substituting the values we found:
\[ f''(c) = \frac{\frac{1}{4} - \frac{1}{2}}{2} = \frac{\frac{1}{4} - \frac{2}{4}}{2} = \frac{-\frac{1}{4}}{2} = -\frac{1}{8} \]

### Final Result
Thus, there exists a number \( c \in (2, 4) \) such that:
\[ f''(c) = -\frac{1}{8} \]