`f(x)={{:(x-1",", -1 le xle 0),(x^(2)",",0 lt x le 1):}` and g(x)=sinx. Find `h(x)=f(abs(g(x)))+abs(f(g(x))).`

Answer

To solve the problem, we need to find \( h(x) = f(\lvert g(x) \rvert) + \lvert f(g(x)) \rvert \) where \( f(x) \) and \( g(x) \) are defined as follows:

- \( f(x) = \begin{cases} x - 1 & \text{if } -1 \leq x \leq 0 \\ x^2 & \text{if } 0 < x \leq 1 \end{cases} \)

- \( g(x) = \sin x \)

### Step-by-Step Solution:

1. **Determine \( g(x) = \sin x \)**:
- The sine function oscillates between -1 and 1 for all real \( x \).

2. **Find \( \lvert g(x) \rvert = \lvert \sin x \rvert \)**:
- Since \( \sin x \) can take negative values, \( \lvert \sin x \rvert \) will always be non-negative and will range from 0 to 1.

3. **Evaluate \( f(\lvert g(x) \rvert) = f(\lvert \sin x \rvert) \)**:
- Since \( \lvert \sin x \rvert \) is always between 0 and 1, we will use the second case of \( f(x) \):
\[ f(\lvert \sin x \rvert) = (\lvert \sin x \rvert)^2 \]

4. **Calculate \( \lvert f(g(x)) \rvert = \lvert f(\sin x) \rvert \)**:
- We need to evaluate \( f(\sin x) \) based on the value of \( \sin x \):
- If \( -1 \leq \sin x \leq 0 \):
\[ f(\sin x) = \sin x - 1 \]
- Here, \( \lvert f(\sin x) \rvert = \lvert \sin x - 1 \rvert = 1 - \sin x \) (since \( \sin x \) is negative or zero).
- If \( 0 < \sin x \leq 1 \):
\[ f(\sin x) = (\sin x)^2 \]
- Here, \( \lvert f(\sin x) \rvert = (\sin x)^2 \).

5. **Combine the results to find \( h(x) \)**:
- For \( -1 \leq \sin x \leq 0 \):
\[ h(x) = (\lvert \sin x \rvert)^2 + (1 - \sin x) = \sin^2 x + 1 - \sin x \]
- For \( 0 < \sin x \leq 1 \):
\[ h(x) = (\lvert \sin x \rvert)^2 + (\sin x)^2 = 2\sin^2 x \]

6. **Final expression for \( h(x) \)**:
- Thus, we can summarize \( h(x) \) as:
\[ h(x) = \begin{cases} \sin^2 x + 1 - \sin x & \text{if } -1 \leq \sin x \leq 0 \\ 2\sin^2 x & \text{if } 0 < \sin x \leq 1 \end{cases} \]