Let f(x) and g(x) be two functions given by `f(x)=-1|x-1|,-1 le x le 3` and `g(x)=2-|x+1|,-2 le x le 2`
Then,

Options

A. fog is differentiable at x=-1 and gof is differentiable at x=1
B. for is differentiable at x=-1 and gof is not differentiable at x=1
C. fog is differentiable at x=1 and gof is differentiable at x=-1
D. none of these

Explanation

To solve the problem, we need to analyze the two functions \( f(x) \) and \( g(x) \), and then find the compositions \( f(g(x)) \) and \( g(f(x)) \).

### Step 1: Define the Functions
The functions are given as:
- \( f(x) = -1 |x - 1| \) for \( -1 \leq x \leq 3 \)
- \( g(x) = 2 - |x + 1| \) for \( -2 \leq x \leq 2 \)

### Step 2: Analyze \( f(x) \)
The function \( f(x) \) can be rewritten based on the definition of the absolute value:
- For \( x < 1 \): \( f(x) = -1(1 - x) = x - 1 \)
- For \( x \geq 1 \): \( f(x) = -1(x - 1) = -x + 1 \)

Thus, we can express \( f(x) \) as:
\[ f(x) = \begin{cases} x - 1 & \text{if } -1 \leq x < 1 \\ -x + 1 & \text{if } 1 \leq x \leq 3 \end{cases} \]

### Step 3: Analyze \( g(x) \)
Similarly, for \( g(x) \):
- For \( x < -1 \): \( g(x) = 2 - (-x - 1) = 2 + x + 1 = x + 3 \)
- For \( -1 \leq x < 1 \): \( g(x) = 2 - (x + 1) = 1 - x \)
- For \( x \geq 1 \): \( g(x) = 2 - (x + 1) = 1 - x \)

Thus, we can express \( g(x) \) as:
\[ g(x) = \begin{cases} x + 3 & \text{if } -2 \leq x < -1 \\ 1 - x & \text{if } -1 \leq x \leq 2 \end{cases} \]

### Step 4: Find \( f(g(x)) \)
Now we will find \( f(g(x)) \):
1. For \( -2 \leq x < -1 \):
- \( g(x) = x + 3 \)
- Since \( x + 3 \) will be greater than 1, we use \( f(x) = -x + 1 \):
\[ f(g(x)) = f(x + 3) = - (x + 3) + 1 = -x - 2 \]

2. For \( -1 \leq x \leq 2 \):
- \( g(x) = 1 - x \)
- Depending on \( x \):
- If \( -1 \leq x < 1 \): \( g(x) \) is greater than 1, so:
\[ f(g(x)) = - (1 - x) + 1 = x \]
- If \( 1 \leq x \leq 2 \): \( g(x) \) is less than or equal to 1, so:
\[ f(g(x)) = - (1 - x) + 1 = x \]

Thus, we can summarize:
\[ f(g(x)) = \begin{cases} -x - 2 & \text{if } -2 \leq x < -1 \\ x & \text{if } -1 \leq x \leq 2 \end{cases} \]

### Step 5: Find \( g(f(x)) \)
Now we will find \( g(f(x)) \):
1. For \( -1 \leq x < 1 \):
- \( f(x) = x - 1 \) (which is less than 0)
- So, \( g(f(x)) = g(x - 1) = 2 - |(x - 1) + 1| = 2 - |x| \)

2. For \( 1 \leq x \leq 3 \):
- \( f(x) = -x + 1 \) (which is less than 0)
- So, \( g(f(x)) = g(-x + 1) = 2 - |(-x + 1) + 1| = 2 - (2 - x) = x \)

Thus, we can summarize:
\[ g(f(x)) = \begin{cases} 2 - |x| & \text{if } -1 \leq x < 1 \\ x & \text{if } 1 \leq x \leq 3 \end{cases} \]

### Conclusion
We have found the compositions \( f(g(x)) \) and \( g(f(x)) \).