If `f(x)=|x-2|` and `g(x)=f(f(x))` then `g'(x)=`

Answer

To solve the problem, we need to find the derivative of the function \( g(x) = f(f(x)) \) where \( f(x) = |x - 2| \).

### Step 1: Define the function \( f(x) \)

The function \( f(x) = |x - 2| \) can be expressed as a piecewise function:

\[ f(x) = \begin{cases} 2 - x & \text{if } x < 2 \\ x - 2 & \text{if } x \geq 2 \end{cases} \]

### Step 2: Determine \( g(x) = f(f(x)) \)

Now we need to find \( g(x) = f(f(x)) \). We will evaluate \( f(f(x)) \) for both cases of \( f(x) \).

#### Case 1: \( x < 2 \)

For \( x < 2 \), we have:

\[ f(x) = 2 - x \]

Now we need to find \( f(f(x)) = f(2 - x) \):

- If \( 2 - x < 2 \) (which is true for \( x < 2 \)):

\[ f(2 - x) = 2 - (2 - x) = x \]

Thus, for \( x < 2 \):

\[ g(x) = f(f(x)) = x \]

#### Case 2: \( x \geq 2 \)

For \( x \geq 2 \), we have:

\[ f(x) = x - 2 \]

Now we need to find \( f(f(x)) = f(x - 2) \):

- If \( x - 2 < 2 \) (which is true for \( 2 \leq x < 4 \)):

\[ f(x - 2) = 2 - (x - 2) = 4 - x \]

- If \( x - 2 \geq 2 \) (which is true for \( x \geq 4 \)):

\[ f(x - 2) = (x - 2) - 2 = x - 4 \]

Thus, for \( 2 \leq x < 4 \):

\[ g(x) = f(f(x)) = 4 - x \]

And for \( x \geq 4 \):

\[ g(x) = f(f(x)) = x - 4 \]

### Step 3: Combine the results for \( g(x) \)

We can summarize \( g(x) \) as follows:

\[ g(x) = \begin{cases} x & \text{if } x < 2 \\ 4 - x & \text{if } 2 \leq x < 4 \\ x - 4 & \text{if } x \geq 4 \end{cases} \]

### Step 4: Find the derivative \( g'(x) \)

Now we will find the derivative \( g'(x) \) for each case:

- For \( x < 2 \):

\[ g'(x) = \frac{d}{dx}(x) = 1 \]

- For \( 2 \leq x < 4 \):

\[ g'(x) = \frac{d}{dx}(4 - x) = -1 \]

- For \( x \geq 4 \):

\[ g'(x) = \frac{d}{dx}(x - 4) = 1 \]

### Final Result

Thus, the derivative \( g'(x) \) can be summarized as:

\[ g'(x) = \begin{cases} 1 & \text{if } x < 2 \\ -1 & \text{if } 2 \leq x < 4 \\ 1 & \text{if } x \geq 4 \end{cases} \]