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Let's go through the problem step by step.

We have:
1. A continuous and differentiable function \( f(x) \).
2. Given condition: 
   \[
   f(2x) = (e^x + 1) f(x) \quad \forall x \in \mathbb{R}
   \]
3. We need to find the value of \( f(0) + f'(0) + f''(0) \).

To solve this, let's start by differentiating both sides of the equation.

### Step 1: Differentiate with respect to \( x \)
Using the given equation:
\[
f(2x) = (e^x + 1) f(x)
\]
we differentiate both sides with respect to \( x \).

Using the chain rule on the left side:
\[
\frac{d}{dx} f(2x) = 2 f'(2x)
\]
For the right side:
\[
\frac{d}{dx} \left((e^x + 1) f(x)\right) = e^x f(x) + (e^x + 1) f'(x)
\]
Thus, we have:
\[
2 f'(2x) = e^x f(x) + (e^x + 1) f'(x)
\]

### Step 2: Substitute \( x = 0 \)
Substituting \( x = 0 \) into the original equation, we get:
\[
f(0) = (e^0 + 1) f(0) \Rightarrow f(0) = 2f(0)
\]
From this, we see that \( f(0) = 0 \).

### Step 3: Use \( f(0) = 0 \) in the Derivatives
Since \( f(0) = 0 \), we can use it to find further values of derivatives at \( x = 0 \). Based on the given solution (D), we are provided

Let \( f(x) \) be a continuous and differentiable function such that \( f(2x) = (e^x + 1) f(x) \forall x \in \mathbb{R} \). Then the value of \( f(0) + f'(0) + f''(0) \) must be?

Learn how to solve the functional equation f(2x) = (e^x + 1)f(x) and find the value of f(0) + f'(0) + f''(0). This problem involves differentiation, exponential functions, and the chain rule, ideal for calculus students.

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