If `A={-2,-1,0,1,2}` and `f : A->B` is an onto function defined by `f(x)=x^2+x+1` and find `B`.

Answer

To find the set \( B \) for the onto function \( f: A \to B \) defined by \( f(x) = x^2 + x + 1 \) where \( A = \{-2, -1, 0, 1, 2\} \), we will calculate the function values for each element in set \( A \).

### Step-by-Step Solution:

1. **Calculate \( f(-2) \)**:
\[ f(-2) = (-2)^2 + (-2) + 1 = 4 - 2 + 1 = 3 \]

2. **Calculate \( f(-1) \)**:
\[ f(-1) = (-1)^2 + (-1) + 1 = 1 - 1 + 1 = 1 \]

3. **Calculate \( f(0) \)**:
\[ f(0) = (0)^2 + (0) + 1 = 0 + 0 + 1 = 1 \]

4. **Calculate \( f(1) \)**:
\[ f(1) = (1)^2 + (1) + 1 = 1 + 1 + 1 = 3 \]

5. **Calculate \( f(2) \)**:
\[ f(2) = (2)^2 + (2) + 1 = 4 + 2 + 1 = 7 \]

6. **Collect the results**:
From the calculations, we have:
- \( f(-2) = 3 \)
- \( f(-1) = 1 \)
- \( f(0) = 1 \)
- \( f(1) = 3 \)
- \( f(2) = 7 \)

7. **Identify the unique outputs**:
The unique outputs from the calculations are \( 1, 3, \) and \( 7 \).

Thus, the set \( B \) is:
\[ B = \{1, 3, 7\} \]