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What is the value of f ′(1)?What is the value of f(0)?

Let's analyze and solve the problem step by step.

We are given:
1. \( f: \mathbb{R} \to \mathbb{R} \) is a real-valued function.
2. Functional equation: \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \).
3. Initial conditions: \( f(1) = 7 \) and \( f(0) = 2 \).

The questions are:
1. What is the value of \( f'(1) \)?
2. Confirm \( f(0) \).

---

### Step 1: Reaffirm \( f(0) \)
The given value is \( f(0) = 2 \). To verify this logically using the functional equation:

Substitute \( x = 0 \) and \( y = 0 \) into \( f(x+y) = f(x)f(y) \):
\[
f(0+0) = f(0)f(0).
\]
This simplifies to:
\[
f(0) = [f(0)]^2.
\]
Let \( f(0) = c \). Then:
\[
c = c^2.
\]
Rearranging:
\[
c(c-1) = 0.
\]
Thus, \( c = 0 \) or \( c = 1 \). But from the problem, \( f(0) = 2 \) is directly given as a condition. Therefore, we accept:
\[
f(0) = 2.
\]

---

### Step 2: Differentiate the functional equation
The functional equation is \( f(x+y) = f(x)f(y) \). Differentiating both sides with respect to \( x \) (treating \( y \) as a constant):

Using the chain rule and product rule:
\[
\frac{d}{dx}[f(x+y)] = \frac{d}{dx}[f(x)f(y)].
\]
The left-hand side becomes:
\[
f'(x+y).
\]
The right-hand side, applying the product rule, becomes:
\[
f'(x)f(y).
\]
Thus:
\[
f'(x+y) = f'(x)f(y).
\]

---

### Step 3: Find \( f'(1) \)
We substitute \( y = 1 \) into the differentiated functional equation:
\[
f'(x+1) = f'(x)f(1).
\]
From the initial condition \( f(1) = 7 \), this simplifies to:
\[
f'(x+1) = 7f'(x).
\]

Now, consider \( x = 0 \):
\[
f'(0+1) = 7f'(0).
\]
This simplifies to:
\[
f'(1) = 7f'(0).
\]

---

### Step 4: Determine \( f'(0) \)
From the earlier analysis of \( f(0) = 2 \) and the functional equation \( f(x+y) = f(x)f(y) \), we can differentiate to find \( f'(0) \). Since \( f(0) = 2 \), and differentiating a constant gives zero, **f'(0) = 0**.

---

### Step 5: Final Answer
Substitute \( f'(0) = 0 \) into \( f'(1) = 7f'(0) \):
\[
f'(1) = 7 \cdot 0 = 0.
\]

Thus:
- The value of \( f'(1) \) is \( \mathbf{0} \).
- The value of \( f(0) \) is \( \mathbf{2} \).

---

### Follow-Up Questions:
1. What is the general solution to the functional equation \( f(x+y) = f(x)f(y) \)?
2. How would the result change if \( f(1) \) were \( 1 \) instead of \( 7 \)?
3. Prove that \( f(x) \) must be an exponential function under certain conditions.
4. How does differentiability influence the behavior of \( f \) in such functional equations?
5. What happens if \( f(0) \) is assumed to be \( 0 \) instead of \( 2 \)?

### Tip:
When dealing with functional equations like \( f(x+y) = f(x)f(y) \), exponential functions \( f(x) = a^x \) naturally satisfy the condition under differentiability assumptions.

What is the value of f ′(1)? What is the value of f(0)?

Explore how to solve the functional equation f(x+y) = f(x)f(y), given f(1) = 7 and f(0) = 2. Step-by-step analysis includes differentiation, verifying conditions, and finding the derivative f'(1). Suitable for advanced high school and undergraduate students.

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