Let `f(x)={(tan^(-1)x, , |x|ge1),((x^(2)-1)/4, , |x|lt1):}` then domain of f'(x) is

Options

A. `(-oo,oo)~{1}`
B. `(-oo, oo)~{-1}`
C. `(-oo,oo)~{1,-1}`
D. `(-oo,oo)~{-1,0,1}`

Answer

To find the domain of \( f'(x) \) for the piecewise function

\[ f(x) = \begin{cases} \tan^{-1}(x) & \text{if } |x| \geq 1 \\ \frac{x^2 - 1}{4} & \text{if } |x| < 1 \end{cases} \]

we need to follow these steps:

### Step 1: Differentiate each piece of the function

1. For \( |x| \geq 1 \), we have \( f(x) = \tan^{-1}(x) \).
- The derivative is:
\[ f'(x) = \frac{1}{1 + x^2} \]

2. For \( |x| < 1 \), we have \( f(x) = \frac{x^2 - 1}{4} \).
- The derivative is:
\[ f'(x) = \frac{2x}{4} = \frac{x}{2} \]

### Step 2: Identify the points of interest

The function \( f(x) \) changes its definition at \( x = -1 \) and \( x = 1 \). We need to check the continuity of \( f(x) \) at these points to determine where \( f'(x) \) is defined.

### Step 3: Check continuity at \( x = 1 \)

- Right-hand limit as \( x \to 1^+ \):
\[ f(1) = \tan^{-1}(1) = \frac{\pi}{4} \]

- Left-hand limit as \( x \to 1^- \):
\[ f(1) = \frac{1^2 - 1}{4} = 0 \]

Since \( \frac{\pi}{4} \neq 0 \), \( f(x) \) is not continuous at \( x = 1 \).

### Step 4: Check continuity at \( x = -1 \)

- Right-hand limit as \( x \to -1^+ \):
\[ f(-1) = \frac{(-1)^2 - 1}{4} = 0 \]

- Left-hand limit as \( x \to -1^- \):
\[ f(-1) = \tan^{-1}(-1) = -\frac{\pi}{4} \]

Since \( 0 \neq -\frac{\pi}{4} \), \( f(x) \) is not continuous at \( x = -1 \).

### Step 5: Determine the domain of \( f'(x) \)

The derivative \( f'(x) \) is defined everywhere except at the points where \( f(x) \) is not continuous. Therefore, the domain of \( f'(x) \) is:

\[ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \]

### Final Answer

The domain of \( f'(x) \) is:

\[ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) \]