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**: We start with the function given in the problem:
\[ f(x) = x^2 + x + 1 \]

2. **Calculate \( f(x-1) \)**: We need to find \( f(x-1) \):
\[ f(x-1) = (x-1)^2 + (x-1) + 1 \]
Expanding this:
\[ = (x^2 - 2x + 1) + (x - 1) + 1 \]
Simplifying further:
\[ = x^2 - 2x + 1 + x - 1 + 1 \]
Combine like terms:
\[ = x^2 - 2x + x + 1 - 1 + 1 = x^2 - x + 1 \]

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

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

5. **Solve for \( x \)**: From the equation \( -x = 0 \), we find:
\[ x = 0 \]

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