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

Explanation

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

### Step 1: Simplify \( f(x) \)
Since \( x > 20 \), we can simplify \( f(x) \):
\[ f(x) = |x - 2| = x - 2 \]
because \( x - 2 \) is positive when \( x > 2 \).

### Step 2: Find \( g(x) = f(f(x)) \)
Now we need to find \( g(x) \):
\[ g(x) = f(f(x)) = f(x - 2) \]
Next, we need to evaluate \( f(x - 2) \):
\[ f(x - 2) = |(x - 2) - 2| = |x - 4| \]
Again, since \( x > 20 \), \( x - 4 \) is also positive:
\[ f(x - 2) = x - 4 \]
Thus, we have:
\[ g(x) = x - 4 \]

### Step 3: Differentiate \( g(x) \)
Now we differentiate \( g(x) \):
\[ g'(x) = \frac{d}{dx}(x - 4) = 1 \]

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