`f(x)={(2+2x,-3lexle0),(1-x/(2),0 lt x le1))`
`g(x)={(-x,-1lexlt0),(x, 0lexle1))`
Range of fog(x) is

Options

A. `[1/2,4]`
B. `[-infty,1/2]cup[1,2]`
C. `[1/2,1]cup{2}`
D. `[1/2,1]cup[4,infty]`

Answer

To find the range of the composite function \( f(g(x)) \), we will follow these steps:

### Step 1: Define the Functions
We have two piecewise functions:
- \( f(x) = \begin{cases} 2 + 2x & \text{for } -3 \leq x \leq 0 \\ 1 - \frac{x}{2} & \text{for } 0 < x \leq 1 \end{cases} \)

- \( g(x) = \begin{cases} -x & \text{for } -1 \leq x < 0 \\ x & \text{for } 0 \leq x \leq 1 \end{cases} \)

### Step 2: Find the Range of \( g(x) \)
First, we need to find the range of \( g(x) \):
- For \( -1 \leq x < 0 \), \( g(x) = -x \) which ranges from \( g(-1) = 1 \) to \( g(0) = 0 \) (not including 0).
- For \( 0 \leq x \leq 1 \), \( g(x) = x \) which ranges from \( g(0) = 0 \) to \( g(1) = 1 \).

Thus, the overall range of \( g(x) \) is \( [0, 1] \).

### Step 3: Evaluate \( f(g(x)) \)
Now we will evaluate \( f(g(x)) \) for the range of \( g(x) \):
1. **For \( g(x) \in [0, 1] \)**:
- When \( g(x) = 0 \), \( f(0) = 2 + 2(0) = 2 \).
- When \( g(x) = 1 \), \( f(1) = 1 - \frac{1}{2} = \frac{1}{2} \).

2. **Evaluate \( f(g(x)) \) for \( g(x) \) values**:
- For \( g(x) \) in the range \( [0, 1] \):
- When \( g(x) \) is in \( [0, 0] \), we use \( f(x) = 2 + 2x \) which gives \( f(g(x)) = 2 + 2(0) = 2 \).
- When \( g(x) \) is in \( (0, 1] \), we use \( f(x) = 1 - \frac{x}{2} \) which gives \( f(g(x)) = 1 - \frac{x}{2} \) where \( x \) varies from \( 0 \) to \( 1 \).

### Step 4: Determine the Range of \( f(g(x)) \)
- For \( g(x) = 0 \), \( f(g(x)) = 2 \).
- For \( g(x) = 1 \), \( f(g(x)) = \frac{1}{2} \).

As \( g(x) \) varies from \( 0 \) to \( 1 \), \( f(g(x)) \) will decrease from \( 2 \) to \( \frac{1}{2} \).

### Step 5: Conclusion
The range of \( f(g(x)) \) is from \( \frac{1}{2} \) to \( 2 \).

### Final Answer
The range of \( f(g(x)) \) is \( \left[\frac{1}{2}, 2\right] \).

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