`f (x)= {{:(2(x+1),,, x le -1),( sqrt(1- x^(2)),,, -1 lt x lt 1), (|||x|-1|-1|,,, x ge1 ):}`, then:

Options

A. f (x) is non-differentialbe at exactly three points
B. f (x) is continous in `(-oo,1]`
C. `f(x)` is differentiable in `(-oo, -1)`
D. f (x) is finite type of discontinuity at `x=1,` but continous at `x=-1`

Answer

To solve the problem, we need to analyze the piecewise function given by:

\[ f(x) = \begin{cases} 2(x + 1) & \text{if } x \leq -1 \\ \sqrt{1 - x^2} & \text{if } -1 < x < 1 \\ ||x| - 1| - 1 & \text{if } x \geq 1 \end{cases} \]

### Step 1: Analyze the first piece \( f(x) = 2(x + 1) \) for \( x \leq -1 \)

For \( x \leq -1 \):
- This is a linear function with a slope of 2.
- At \( x = -1 \), \( f(-1) = 2(-1 + 1) = 0 \).

### Step 2: Analyze the second piece \( f(x) = \sqrt{1 - x^2} \) for \( -1 < x < 1 \)

For \( -1 < x < 1 \):
- This is the upper half of a circle with radius 1 centered at the origin.
- At \( x = -1 \), \( f(-1) = \sqrt{1 - (-1)^2} = 0 \).
- At \( x = 1 \), \( f(1) = \sqrt{1 - 1^2} = 0 \).

### Step 3: Analyze the third piece \( f(x) = ||x| - 1| - 1 \) for \( x \geq 1 \)

For \( x \geq 1 \):
- Since \( x \) is positive, \( |x| = x \).
- Thus, \( ||x| - 1| - 1 = |x - 1| - 1 = x - 1 - 1 = x - 2 \) (for \( x \geq 1 \)).
- At \( x = 1 \), \( f(1) = 1 - 2 = -1 \).

### Step 4: Check continuity at the transition points \( x = -1 \) and \( x = 1 \)

1. **At \( x = -1 \)**:
- Left-hand limit: \( \lim_{x \to -1^-} f(x) = 0 \)
- Right-hand limit: \( \lim_{x \to -1^+} f(x) = \sqrt{1 - (-1)^2} = 0 \)
- Since both limits are equal and \( f(-1) = 0 \), \( f(x) \) is continuous at \( x = -1 \).

2. **At \( x = 1 \)**:
- Left-hand limit: \( \lim_{x \to 1^-} f(x) = \sqrt{1 - 1^2} = 0 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = 1 - 2 = -1 \)
- Since the left-hand limit and right-hand limit are not equal, \( f(x) \) is discontinuous at \( x = 1 \).

### Step 5: Check differentiability at the transition points

1. **At \( x = -1 \)**:
- Left-hand derivative: \( f'(x) = 2 \) (for \( x \leq -1 \))
- Right-hand derivative:
\[ \lim_{h \to 0^+} \frac{f(-1 + h) - f(-1)}{h} = \lim_{h \to 0^+} \frac{\sqrt{1 - (-1 + h)^2} - 0}{h} \]
This limit approaches 0 as \( h \to 0^+ \).
- Since the left-hand derivative (2) and right-hand derivative (0) are not equal, \( f(x) \) is not differentiable at \( x = -1 \).

2. **At \( x = 1 \)**:
- Left-hand derivative: \( f'(x) = 0 \) (for \( -1 < x < 1 \))
- Right-hand derivative:
\[ \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h} = \lim_{h \to 0^+} \frac{(1 + h) - 2 - (-1)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]
- Since the left-hand derivative (0) and right-hand derivative (1) are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

### Final Conclusion

- The function \( f(x) \) is non-differentiable at exactly two points: \( x = -1 \) and \( x = 1 \).
- It is continuous at \( x = -1 \) but discontinuous at \( x = 1 \).