Find the domain and range of real function ,`f(x)=sqrt(x^2 +4)`

Answer

To find the domain and range of the function \( f(x) = \sqrt{x^2 + 4} \), we will follow these steps:

### Step 1: Determine the Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined.

1. **Identify the expression under the square root**:
\[ x^2 + 4 \]
Since \( x^2 \) is always non-negative (i.e., \( x^2 \geq 0 \)), adding 4 ensures that the expression \( x^2 + 4 \) is always positive.

2. **Conclusion about the domain**:
Since \( x^2 + 4 \) is always positive for all real numbers \( x \), the function \( f(x) \) is defined for all real numbers.
\[ \text{Domain} = \mathbb{R} \]

### Step 2: Determine the Range
The range of a function is the set of all possible output values (y-values) that the function can produce.

1. **Set the function equal to y**:
\[ y = \sqrt{x^2 + 4} \]

2. **Square both sides**:
\[ y^2 = x^2 + 4 \]
Rearranging gives:
\[ x^2 = y^2 - 4 \]

3. **Determine when the right side is non-negative**:
For \( x^2 \) to be non-negative, we need:
\[ y^2 - 4 \geq 0 \]
This simplifies to:
\[ y^2 \geq 4 \]

4. **Solve for y**:
Taking the square root of both sides gives:
\[ |y| \geq 2 \]
This means:
\[ y \leq -2 \quad \text{or} \quad y \geq 2 \]

5. **Consider the nature of the function**:
Since \( f(x) = \sqrt{x^2 + 4} \) is always non-negative (as square roots yield non-negative results), we discard the negative part of the range:
\[ y \geq 2 \]

6. **Conclusion about the range**:
Therefore, the range of the function is:
\[ \text{Range} = [2, \infty) \]

### Final Answer
- **Domain**: \( \mathbb{R} \)
- **Range**: \( [2, \infty) \)