If `f(x)={ 1+x^2 ; x<=1 ,x+1; 1< x<=2` and `g(x)=1-x ; -2<=x<=1` then define the function `fog(x)`

Explanation

To define the function \( f \circ g(x) \), we start by analyzing the given functions \( f(x) \) and \( g(x) \).

1. **Define the functions:**
- \( f(x) = \begin{cases} 1 + x^2 & \text{if } x \leq 1 \\ x + 1 & \text{if } 1 < x \leq 2 \end{cases} \)
- \( g(x) = 1 - x \) for \( -2 \leq x \leq 1 \)

2. **Find the range of \( g(x) \):**
- Since \( g(x) = 1 - x \), we need to determine the values of \( g(x) \) for \( x \) in the interval \( [-2, 1] \).
- When \( x = -2 \), \( g(-2) = 1 - (-2) = 3 \).
- When \( x = 1 \), \( g(1) = 1 - 1 = 0 \).
- Thus, the range of \( g(x) \) is \( [0, 3] \).

3. **Determine the intervals for \( f(g(x)) \):**
- We need to evaluate \( f(g(x)) \) based on the output of \( g(x) \):
- For \( g(x) \leq 1 \): This corresponds to \( g(x) = 1 - x \leq 1 \) or \( x \geq 0 \).
- For \( 1 < g(x) \leq 2 \): This corresponds to \( 1 < 1 - x \leq 2 \) or \( -1 < x < 0 \).

4. **Evaluate \( f(g(x)) \):**
- **Case 1:** For \( x \geq 0 \) (where \( g(x) \leq 1 \)):
- \( f(g(x)) = f(1 - x) = 1 + (1 - x)^2 \)
- Simplifying:
\[ f(g(x)) = 1 + (1 - 2x + x^2) = 2 - 2x + x^2 \]
- **Case 2:** For \( -1 < x < 0 \) (where \( 1 < g(x) \leq 2 \)):
- \( f(g(x)) = f(1 - x) = (1 - x) + 1 = 2 - x \)

5. **Combine the results:**
- Therefore, we can write the function \( f \circ g(x) \) as:
\[ f \circ g(x) = \begin{cases} 2 - 2x + x^2 & \text{if } x \geq 0 \\ 2 - x & \text{if } -1 < x < 0 \end{cases} \]

### Final Definition of \( f \circ g(x) \):
\[ f \circ g(x) = \begin{cases} 2 - 2x + x^2 & \text{if } x \geq 0 \\ 2 - x & \text{if } -1 < x < 0 \end{cases} \]