If `f(x)={x^3,if x^2<1 and x, if x^2>=1` then `f(x)` is differentiable at (a) `(-oo,oo)-{1}` (b) `(-oo,oo)-{1,-1}` (c) `(-oo,oo)-{1,-1,0}` (d) `(-oo,oo)-{-1}`

Options

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

Answer

To determine where the function \( f(x) \) is differentiable, we first need to analyze the function given by:

\[ f(x) = \begin{cases} x^3 & \text{if } x^2 < 1 \\ x & \text{if } x^2 \geq 1 \end{cases} \]

### Step 1: Identify the intervals
The condition \( x^2 < 1 \) implies that \( -1 < x < 1 \). The condition \( x^2 \geq 1 \) implies that \( x \leq -1 \) or \( x \geq 1 \). Therefore, we can rewrite the function as:

\[ f(x) = \begin{cases} x^3 & \text{if } -1 < x < 1 \\ x & \text{if } x \leq -1 \text{ or } x \geq 1 \end{cases} \]

### Step 2: Check continuity at the boundary points
To check differentiability, we first need to ensure that the function is continuous at the boundary points \( x = -1 \) and \( x = 1 \).

1. **At \( x = -1 \)**:
- From the left: \( \lim_{x \to -1^-} f(x) = (-1)^3 = -1 \)
- From the right: \( \lim_{x \to -1^+} f(x) = -1 \)
- \( f(-1) = -1 \)
- Since both limits and the function value are equal, \( f(x) \) is continuous at \( x = -1 \).

2. **At \( x = 1 \)**:
- From the left: \( \lim_{x \to 1^-} f(x) = (1)^3 = 1 \)
- From the right: \( \lim_{x \to 1^+} f(x) = 1 \)
- \( f(1) = 1 \)
- Since both limits and the function value are equal, \( f(x) \) is continuous at \( x = 1 \).

### Step 3: Check differentiability at the boundary points
Next, we need to check if \( f(x) \) is differentiable at \( x = -1 \) and \( x = 1 \).

1. **At \( x = -1 \)**:
- The derivative from the left: \( f'(x) = 3x^2 \) gives \( f'(-1) = 3(-1)^2 = 3 \)
- The derivative from the right: \( f'(x) = 1 \) gives \( f'(-1) = 1 \)
- Since the left-hand derivative (3) does not equal the right-hand derivative (1), \( f(x) \) is not differentiable at \( x = -1 \).

2. **At \( x = 1 \)**:
- The derivative from the left: \( f'(x) = 3x^2 \) gives \( f'(1) = 3(1)^2 = 3 \)
- The derivative from the right: \( f'(x) = 1 \) gives \( f'(1) = 1 \)
- Since the left-hand derivative (3) does not equal the right-hand derivative (1), \( f(x) \) is not differentiable at \( x = 1 \).

### Step 4: Conclusion
Since \( f(x) \) is not differentiable at \( x = -1 \) and \( x = 1 \), the function is differentiable everywhere else in the real numbers except these points. Therefore, the correct answer is:

\[ (-\infty, \infty) - \{-1, 1\} \]

### Final Answer
(b) \( (-\infty, \infty) - \{-1, 1\} \)