If `f(x)=|x-1|" and "g(x)=f(f(f(x)))`, then for `xgt2,g'(x)` is equal to

Options

A. 1 for `x gt 2`
B. 1 for ` 2 lt x lt 3`
C. `-1` for ` 2 lt x lt 3`
D. `-1` for `x gt 3`

Answer

To solve the problem, we need to find the derivative \( g'(x) \) for the function \( g(x) = f(f(f(x))) \) where \( f(x) = |x - 1| \) and \( x > 2 \).

### Step-by-Step Solution

1. **Define the function \( f(x) \)**:
Since \( x > 2 \), we can simplify \( f(x) \):
\[ f(x) = x - 1 \]

2. **Find \( f(f(x)) \)**:
Now we calculate \( f(f(x)) \):
\[ f(f(x)) = f(x - 1) = (x - 1) - 1 = x - 2 \]

3. **Find \( f(f(f(x))) \)**:
Next, we calculate \( f(f(f(x))) \):
\[ f(f(f(x))) = f(x - 2) = (x - 2) - 1 = x - 3 \]

4. **Define \( g(x) \)**:
Now we can express \( g(x) \):
\[ g(x) = f(f(f(x))) = x - 3 \]

5. **Differentiate \( g(x) \)**:
We find the derivative \( g'(x) \):
\[ g'(x) = \frac{d}{dx}(x - 3) = 1 \]

6. **Conclusion**:
Therefore, for \( x > 2 \), we have:
\[ g'(x) = 1 \]

### Summary of Results
- For \( x > 2 \), \( g'(x) = 1 \).