If `f(x)=|x-2|" and "g(x)=f(f(x)),` then for `xgt20,g'(x)` equals

Options

A. -1
B. 1
C. 0
D. none of these

Explanation

To solve the problem, we need to find \( g'(x) \) given that \( f(x) = |x - 2| \) and \( g(x) = f(f(x)) \) for \( x > 20 \).

### Step-by-step Solution:

1. **Define the function \( f(x) \)**:
Since \( x > 20 \), we can remove the absolute value:
\[ f(x) = |x - 2| = x - 2 \]

2. **Find \( f(f(x)) \)**:
Now we need to find \( g(x) = f(f(x)) \):
\[ f(x) = x - 2 \quad \text{(from step 1)} \]
Now substitute \( f(x) \) into itself:
\[ f(f(x)) = f(x - 2) \]

3. **Evaluate \( f(x - 2) \)**:
Since \( x - 2 > 20 - 2 = 18 \), we again remove the absolute value:
\[ f(x - 2) = |(x - 2) - 2| = |x - 4| = x - 4 \quad \text{(for } x > 20\text{)} \]

4. **Define \( g(x) \)**:
Now we can write:
\[ g(x) = f(f(x)) = x - 4 \]

5. **Differentiate \( g(x) \)**:
Now we differentiate \( g(x) \) with respect to \( x \):
\[ g'(x) = \frac{d}{dx}(x - 4) = 1 \]

### Final Answer:
Thus, for \( x > 20 \), we have:
\[ g'(x) = 1 \]