Let f(x) `= {{:(x + 1 , :, x £ 0) , (x , : , x gt 0):}` and h (x) = |f(x) + f(|x|) ` . Then h(x) is defined by :

Answer

To define the function \( h(x) \) based on the given piecewise function \( f(x) \), we will analyze the function in different cases based on the value of \( x \).

### Step-by-step Solution:

1. **Define the function \( f(x) \)**:
\[ f(x) = \begin{cases} x + 1 & \text{if } x \leq 0 \\ x & \text{if } x > 0 \end{cases} \]

2. **Define \( h(x) = |f(x) + f(|x|)| \)**:
We will analyze \( h(x) \) for three cases: \( x < 0 \), \( x = 0 \), and \( x > 0 \).

3. **Case 1: \( x < 0 \)**:
- Here, \( f(x) = x + 1 \) (since \( x \leq 0 \)).
- Also, \( |x| = -x \) (which is positive), so \( f(|x|) = f(-x) = -x \) (since \(-x > 0\)).
- Therefore, we have:
\[ h(x) = |(x + 1) + (-x)| = |1| = 1 \]

4. **Case 2: \( x = 0 \)**:
- Here, \( f(0) = 0 + 1 = 1 \).
- Also, \( f(|0|) = f(0) = 1 \).
- Therefore, we have:
\[ h(0) = |1 + 1| = |2| = 2 \]

5. **Case 3: \( x > 0 \)**:
- Here, \( f(x) = x \) (since \( x > 0 \)).
- Also, \( |x| = x \), so \( f(|x|) = f(x) = x \).
- Therefore, we have:
\[ h(x) = |x + x| = |2x| = 2x \]

### Final Definition of \( h(x) \):
Combining all the cases, we can define \( h(x) \) as follows:
\[ h(x) = \begin{cases} 1 & \text{if } x < 0 \\ 2 & \text{if } x = 0 \\ 2x & \text{if } x > 0 \end{cases} \]