If `f(x)=x^2+x+1`, then find the value of 'x' for which `f(x-1) =f(x)`

Answer

To solve the problem, we need to find the value of \( x \) for which \( f(x-1) = f(x) \), given that \( f(x) = x^2 + x + 1 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = x^2 + x + 1 \]

2. **Find \( f(x-1) \)**:
Substitute \( x-1 \) into the function:
\[ f(x-1) = (x-1)^2 + (x-1) + 1 \]

3. **Expand \( f(x-1) \)**:
\[ f(x-1) = (x^2 - 2x + 1) + (x - 1) + 1 \]
Combine like terms:
\[ f(x-1) = x^2 - 2x + 1 + x - 1 + 1 = x^2 - x + 1 \]

4. **Set up the equation**:
Now, we need to set \( f(x-1) \) equal to \( f(x) \):
\[ f(x-1) = f(x) \implies x^2 - x + 1 = x^2 + x + 1 \]

5. **Simplify the equation**:
Subtract \( x^2 + 1 \) from both sides:
\[ x^2 - x + 1 - (x^2 + x + 1) = 0 \]
This simplifies to:
\[ -x - x = 0 \implies -2x = 0 \]

6. **Solve for \( x \)**:
Divide both sides by -2:
\[ x = 0 \]

### Final Answer:
The value of \( x \) for which \( f(x-1) = f(x) \) is:
\[ \boxed{0} \]