Let f(x) =|x| and g(x)=|x| where [.] denotes the greatest function. Then, (fog)' (-2) is

Options

A. 0
B. 1
C. `-1`
D. non-existent

Explanation

To solve the problem, we need to analyze the functions \( f(x) = |x| \) and \( g(x) = |x| \), and then find the derivative of the composition \( (f \circ g)'(-2) \).

### Step 1: Understand the functions
We have:
- \( f(x) = |x| \)
- \( g(x) = |x| \)

### Step 2: Find the composition \( f(g(x)) \)
Since \( g(x) = |x| \), we can substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f(|x|) = ||x|| \]
Since \( ||x| = |x| \), we have:
\[ f(g(x)) = |x| \]

### Step 3: Analyze the function \( f(g(x)) \)
The function \( f(g(x)) = |x| \) is defined for all \( x \). However, we need to check its continuity and differentiability at \( x = -2 \).

### Step 4: Check continuity at \( x = -2 \)
The function \( |x| \) is continuous everywhere, including at \( x = -2 \). Thus, \( f(g(x)) \) is continuous at \( x = -2 \).

### Step 5: Check differentiability at \( x = -2 \)
To check differentiability, we need to find the derivative of \( f(g(x)) = |x| \):
\[ \frac{d}{dx} |x| = \begin{cases} 1 & \text{if } x > 0 \\ -1 & \text{if } x < 0 \\ \text{undefined} & \text{if } x = 0 \end{cases} \]
At \( x = -2 \), we have \( x < 0 \), so:
\[ \frac{d}{dx} |x| = -1 \]

### Step 6: Conclusion
Since \( f(g(x)) \) is continuous and differentiable at \( x = -2 \), we can find the derivative:
\[ (f \circ g)'(-2) = -1 \]

### Final Answer
\[ (f \circ g)'(-2) = -1 \]