If `f((x+2y)/3)=(f(x)+2f(y))/3AAx ,y in Ra n df^(prime)(0)=1,f(0)=2,` then find `f(x)dot`

Explanation

To solve the problem, we start with the given functional equation:

\[ f\left(\frac{x + 2y}{3}\right) = \frac{f(x) + 2f(y)}{3} \]

We also know that \( f'(0) = 1 \) and \( f(0) = 2 \). We need to find the function \( f(x) \).

### Step 1: Analyze the Functional Equation

The functional equation resembles the section formula, which suggests that \( f(x) \) could be a linear function. We assume a linear form for \( f(x) \):

\[ f(x) = ax + b \]

### Step 2: Substitute into the Functional Equation

Substituting \( f(x) = ax + b \) into the functional equation:

\[ f\left(\frac{x + 2y}{3}\right) = a\left(\frac{x + 2y}{3}\right) + b = \frac{ax + 2ay}{3} + b \]

Now, we compute the right-hand side:

\[ \frac{f(x) + 2f(y)}{3} = \frac{(ax + b) + 2(ay + b)}{3} = \frac{ax + b + 2ay + 2b}{3} = \frac{ax + 2ay + 3b}{3} \]

### Step 3: Set Both Sides Equal

Now we set both sides of the equation equal to each other:

\[ \frac{ax + 2ay}{3} + b = \frac{ax + 2ay + 3b}{3} \]

### Step 4: Simplify the Equation

Multiplying through by 3 to eliminate the denominator gives:

\[ ax + 2ay + 3b = ax + 2ay + 3b \]

This equation holds true for all \( x \) and \( y \), confirming that our assumption of \( f(x) \) being linear is valid.

### Step 5: Use Given Conditions

We need to determine the constants \( a \) and \( b \) using the conditions \( f'(0) = 1 \) and \( f(0) = 2 \).

From the linear function \( f(x) = ax + b \):

1. **Finding \( b \)**:
\[ f(0) = b = 2 \]

2. **Finding \( a \)**:
The derivative \( f'(x) = a \). Given \( f'(0) = 1 \):
\[ a = 1 \]

### Step 6: Write the Final Function

Now substituting \( a \) and \( b \) back into the linear function:

\[ f(x) = 1 \cdot x + 2 = x + 2 \]

### Conclusion

Thus, the required function is:

\[ f(x) = x + 2 \]