If `f_1 (x) and f_2 (x) `are defined on domains `D_1 and D_2` respectively, then `f_1 (x) + f_2 (x)` is defined on `D_1 uu D_2`.

Answer

To determine whether the statement "If \( f_1(x) \) and \( f_2(x) \) are defined on domains \( D_1 \) and \( D_2 \) respectively, then \( f_1(x) + f_2(x) \) is defined on \( D_1 \cup D_2 \)" is true or false, we can analyze it step by step.

### Step-by-Step Solution:

1. **Define the Functions and Their Domains**:
Let:
\[ f_1(x) = \frac{1}{x - a} \]
The domain \( D_1 \) of \( f_1(x) \) is:
\[ D_1 = \mathbb{R} \setminus \{a\} \]
This means \( f_1(x) \) is defined for all real numbers except \( a \).

Similarly, let:
\[ f_2(x) = \frac{1}{x - b} \]
The domain \( D_2 \) of \( f_2(x) \) is:
\[ D_2 = \mathbb{R} \setminus \{b\} \]
This means \( f_2(x) \) is defined for all real numbers except \( b \).

2. **Find the Sum of the Functions**:
Now, we calculate the sum:
\[ f_1(x) + f_2(x) = \frac{1}{x - a} + \frac{1}{x - b} \]

3. **Determine the Domain of the Sum**:
The sum \( f_1(x) + f_2(x) \) will be defined wherever both \( f_1(x) \) and \( f_2(x) \) are defined. Therefore, the domain of the sum \( D_{sum} \) is:
\[ D_{sum} = \mathbb{R} \setminus \{a, b\} \]
This means \( f_1(x) + f_2(x) \) is defined for all real numbers except \( a \) and \( b \).

4. **Find the Union of the Domains**:
Now, we find the union of the domains \( D_1 \) and \( D_2 \):
\[ D_1 \cup D_2 = (\mathbb{R} \setminus \{a\}) \cup (\mathbb{R} \setminus \{b\}) = \mathbb{R} \setminus \{a, b\} \]
This means that \( D_1 \cup D_2 \) also excludes both \( a \) and \( b \).

5. **Compare the Domains**:
We see that:
\[ D_{sum} = \mathbb{R} \setminus \{a, b\} \]
and
\[ D_1 \cup D_2 = \mathbb{R} \setminus \{a, b\} \]
Thus, the statement is true in this case.

### Conclusion:
The statement "If \( f_1(x) \) and \( f_2(x) \) are defined on domains \( D_1 \) and \( D_2 \) respectively, then \( f_1(x) + f_2(x) \) is defined on \( D_1 \cup D_2 \)" is **true** for this example.