Let `f (x)= [{:(x+1,,, x lt0),((x-1),,, x ge0):}and g (x)=[{:(x+1,,, x lt 0),((x-1)^(2),,, x ge0):}` then the number of points where `g (f(x))` is not differentiable.

Options

A. 0
B. 1
C. 2
D. None of these

Answer

To solve the problem, we need to find the points where the function \( g(f(x)) \) is not differentiable. We start by analyzing the functions \( f(x) \) and \( g(x) \).

### Step 1: Define the functions

The functions are defined as follows:

\[ f(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ x - 1 & \text{if } x \geq 0 \end{cases} \]

\[ g(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ (x - 1)^2 & \text{if } x \geq 0 \end{cases} \]

### Step 2: Determine the range of \( f(x) \)

1. For \( x < 0 \):
- \( f(x) = x + 1 \) which approaches 1 as \( x \) approaches 0 from the left.

2. For \( x \geq 0 \):
- \( f(x) = x - 1 \) which is 0 at \( x = 1 \) and increases thereafter.

Thus, the range of \( f(x) \) is \( (-\infty, 1) \) for \( x < 0 \) and \( [0, \infty) \) for \( x \geq 0 \).

### Step 3: Find \( g(f(x)) \)

Now we need to find \( g(f(x)) \) for different intervals of \( x \):

1. **For \( x < -1 \)**:
- Here, \( f(x) = x + 1 < 0 \).
- Thus, \( g(f(x)) = g(x + 1) = (x + 1) + 1 = x + 2 \).

2. **For \( -1 \leq x < 0 \)**:
- Here, \( f(x) = x + 1 \) which is between 0 and 1.
- Thus, \( g(f(x)) = g(x + 1) = (x + 1 - 1)^2 = x^2 \).

3. **For \( 0 \leq x < 1 \)**:
- Here, \( f(x) = x - 1 < 0 \).
- Thus, \( g(f(x)) = g(x - 1) = (x - 1) + 1 = x \).

4. **For \( x \geq 1 \)**:
- Here, \( f(x) = x - 1 \geq 0 \).
- Thus, \( g(f(x)) = g(x - 1) = (x - 1)^2 \).

### Step 4: Combine the results

Now we can summarize \( g(f(x)) \):

\[ g(f(x)) = \begin{cases} x + 2 & \text{if } x < -1 \\ x^2 & \text{if } -1 \leq x < 0 \\ x & \text{if } 0 \leq x < 1 \\ (x - 1)^2 & \text{if } x \geq 1 \end{cases} \]

### Step 5: Identify points of non-differentiability

To find points where \( g(f(x)) \) is not differentiable, we check the points where the definition of \( g(f(x)) \) changes:

1. At \( x = -1 \):
- Left limit: \( g(f(-1)) = (-1)^2 = 1 \)
- Right limit: \( g(f(-1)) = -1 + 2 = 1 \)
- The derivatives from both sides are different.

2. At \( x = 0 \):
- Left limit: \( g(f(0^-)) = 0^2 = 0 \)
- Right limit: \( g(f(0^+)) = 0 \)
- The derivatives from both sides are different.

3. At \( x = 1 \):
- Left limit: \( g(f(1^-)) = 1 \)
- Right limit: \( g(f(1^+)) = (1 - 1)^2 = 0 \)
- The derivatives from both sides are different.

### Conclusion

Thus, \( g(f(x)) \) is not differentiable at three points: \( x = -1, 0, 1 \).

The final answer is that there are **3 points** where \( g(f(x)) \) is not differentiable.

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