Consider the following for the next two items that follow :
Let `f(x)=[|x|-|x-1|]^(2)`
What is f'(x) equal to when `0ltxlt1` ?

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 \( 0 < x < 1 \), we can follow these steps:

### Step 1: Define the function for the interval \( 0 < x < 1 \)
In the interval \( 0 < x < 1 \):
- \( |x| = x \) (since \( x \) is positive)
- \( |x - 1| = 1 - x \) (since \( x < 1 \))

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

### Step 2: Differentiate the function
Now we need to differentiate \( f(x) = (2x - 1)^2 \).

Using the chain rule:
\[ f'(x) = 2(2x - 1) \cdot \frac{d}{dx}(2x - 1) \]

The derivative of \( 2x - 1 \) is \( 2 \), so:
\[ f'(x) = 2(2x - 1) \cdot 2 = 4(2x - 1) \]

### Step 3: Simplify the derivative
Now we can simplify the expression:
\[ f'(x) = 8x - 4 \]

### Final Answer
Thus, the derivative \( f'(x) \) when \( 0 < x < 1 \) is:
\[ f'(x) = 8x - 4 \]

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