Consider the following for the next two items that follow :
Let `f(x)=[|x|-|x-1|]^(2)`
What is f'(x) equal to when `xgt1` ?

Options

A. A) 0
B. B) `2x-1`
C. C) `4x-2`
D. D) `8x-4`

Answer

To find \( f'(x) \) for the function \( f(x) = [|x| - |x-1|]^2 \) when \( x > 1 \), we can follow these steps:

### Step 1: Analyze the function for \( x > 1 \)

When \( x > 1 \):
- \( |x| = x \) (since \( x \) is positive)
- \( |x - 1| = x - 1 \) (since \( x - 1 \) is also positive)

Thus, we can rewrite the function:
\[ f(x) = [x - (x - 1)]^2 \]

### Step 2: Simplify the expression

Now simplify the expression inside the brackets:
\[ f(x) = [x - x + 1]^2 = [1]^2 = 1 \]

### Step 3: Determine the derivative

Since \( f(x) = 1 \) is a constant function for \( x > 1 \), the derivative of a constant is:
\[ f'(x) = 0 \]

### Conclusion

Therefore, when \( x > 1 \):
\[ f'(x) = 0 \]