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We are given that \( f: \mathbb{R} \to \mathbb{R} \) is a function satisfying the functional equation

\[
f(x + y) = f(x) + f(y) \quad \text{for all} \, x, y \in \mathbb{R}.
\]

Additionally, we are told that \( f \) is differentiable at \( x = 0 \), with \( f'(0) = a \), and we are asked to show that \( f(x) = ax \).

### Step 1: Consider the form of the functional equation

From the functional equation \( f(x + y) = f(x) + f(y) \), we recognize that this is a well-known Cauchy functional equation. The general solution to this equation is that \( f(x) = cx \) for some constant \( c \), provided that \( f \) is continuous (or differentiable, as in this case). We will now show this step by step.

### Step 2: Use the differentiability condition at \( x = 0 \)

Since \( f \) is differentiable at \( x = 0 \), we can use the definition of the derivative:

\[
f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}.
\]

Given that \( f(0 + 0) = f(0) + f(0) \), it follows that \( f(0) = 0 \). Hence, the derivative at 0 becomes:

\[
a = f'(0) = \lim_{h \to 0} \frac{f(h)}{h}.
\]

This tells us that \( \lim_{h \to 0} \frac{f(h)}{h} = a \), which suggests that for small values of \( h \), \( f(h) \approx ah \). 

### Step 3: Generalizing the result for all \( x \)

Since the functional equation is valid for all real \( x \), we conjecture that \( f(x) = ax \) for all \( x \in \mathbb{R} \). To prove this rigorously, let’s use the fact that \( f \) satisfies the Cauchy functional equation, and it is differentiable at \( x = 0 \).

The general solution to the Cauchy functional equation \( f(x + y) = f(x) + f(y) \), under the assumption of differentiability (which implies continuity), is of the form \( f(x) = cx \), where \( c \) is some constant. From Step 2, we know that \( f'(0) = a \), which implies that \( c = a \).

Thus, \( f(x) = ax \) for all \( x \in \mathbb{R} \).

### Conclusion

We have shown that the function \( f(x) = ax \) satisfies both the functional equation and the differentiability condition at \( x = 0 \) with \( f'(0) = a \). Therefore, the solution to the problem is:

\[
f(x) = ax.
\]

Let me know if you would like any additional clarifications.

### Related Questions:

1. What is the general form of solutions to Cauchy's functional equation without assuming continuity?
2. How can we prove that differentiability implies continuity in the context of this functional equation?
3. Could the function \( f(x) = ax \) still be valid if \( f \) was not differentiable but only continuous?
4. How would the solution change if \( f'(0) \) was not given as \( a \)?
5. Can the assumption that \( f \) is differentiable at just one point be weakened?

### Tip: 
When dealing with functional equations, always check if it's a known form (like Cauchy's equation) as this often leads directly to the solution.

Let f: R → R be a function satisfying f(x + y) = f(x) + f(y) for all x, y ∈ R. Suppose f is differentiable at x = 0, with f'(0) = a. Show that f(x) = ax.

Explore how to solve a functional equation involving differentiability and Cauchy's equation. Learn the step-by-step process of proving that f(x) = ax under the given conditions.

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