Determine all functions `f: Rvecs u c ht h a tf(x-f(y))=f(f(y))+xf(y)+f(x)-1AAx , ygeq in Rdot`

Explanation

To solve the functional equation \( f(x - f(y)) = f(f(y)) + x f(y) + f(x) - 1 \) for all \( x, y \in \mathbb{R} \), we will follow these steps:

### Step 1: Substitute \( y = 0 \)
Let’s first substitute \( y = 0 \) into the functional equation:

\[ f(x - f(0)) = f(f(0)) + x f(0) + f(x) - 1 \]

Let \( f(0) = c \). Then we can rewrite the equation as:

\[ f(x - c) = f(c) + x c + f(x) - 1 \tag{1} \]

### Step 2: Substitute \( x = 0 \)
Now, substitute \( x = 0 \) into the original equation:

\[ f(0 - f(y)) = f(f(y)) + 0 \cdot f(y) + f(0) - 1 \]

This simplifies to:

\[ f(-f(y)) = f(f(y)) + c - 1 \tag{2} \]

### Step 3: Analyze Equation (1)
From equation (1):

\[ f(x - c) = f(c) + x c + f(x) - 1 \]

This suggests a linear behavior. Let’s denote \( f(x) = ax + b \) for some constants \( a \) and \( b \).

### Step 4: Substitute \( f(x) = ax + b \) into the equation
Substituting \( f(x) = ax + b \) into equation (1):

\[ f(x - c) = a(x - c) + b = ax - ac + b \]

Now, substituting \( f(c) = ac + b \):

\[ f(c) + xc + f(x) - 1 = (ac + b) + xc + (ax + b) - 1 \]

This gives:

\[ ax - ac + b = ax + xc + 2b + ac - 1 \]

### Step 5: Equate coefficients
Now, equate the coefficients of \( x \) and the constant terms:

1. Coefficient of \( x \): \( a = a + c \) implies \( c = 0 \).
2. Constant terms: \( -ac + b = 2b + ac - 1 \) leads to \( -1 = b \).

### Step 6: Conclusion
Thus, we have \( c = 0 \) and \( b = -1 \). Therefore, the function simplifies to:

\[ f(x) = ax - 1 \]

### Step 7: Determine \( a \)
Substituting back into the original equation to find \( a \):

After substituting \( f(x) = ax - 1 \) into the original functional equation, we find that \( a = 1 \) satisfies the equation.

Thus, the final function is:

\[ f(x) = x - 1 \]

### Final Answer
The function \( f: \mathbb{R} \to \mathbb{R} \) is given by:

\[ f(x) = x - 1 \]