If `f(x+f(y))=f(x)+yAAx ,y in Ra n df(0)=1,` then find the value of `f(7)dot`

Explanation

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

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

and the condition:

\[ f(0) = 1. \]

We need to find the value of \( f(7) \).

### Step 1: Substitute \( y = 0 \)

Let's substitute \( y = 0 \) into the functional equation:

\[ f(x + f(0)) = f(x) + 0. \]

Since \( f(0) = 1 \), we have:

\[ f(x + 1) = f(x). \]

### Step 2: Identify the periodicity of the function

The equation \( f(x + 1) = f(x) \) shows that the function \( f \) is periodic with a period of 1. This means that the function takes the same value for any two inputs that differ by an integer.

### Step 3: Find \( f(1) \)

Since \( f(0) = 1 \) and \( f \) is periodic with period 1, we have:

\[ f(1) = f(0) = 1. \]

### Step 4: Find \( f(2) \)

Using the periodicity again:

\[ f(2) = f(1) = 1. \]

### Step 5: Generalize for any integer

Continuing this process, we can see that:

\[ f(3) = f(2) = 1, \]
\[ f(4) = f(3) = 1, \]
\[ f(5) = f(4) = 1, \]
\[ f(6) = f(5) = 1, \]
\[ f(7) = f(6) = 1. \]

Thus, we conclude that:

\[ f(7) = 1. \]

### Final Answer

The value of \( f(7) \) is:

\[ \boxed{1}. \]

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