What is the domain of the function `f(x) = sqrt(x^(2)+3)` ?

Options

A. `-1.73 le x le 1.73`
B. `-1.32 le x le 1.32`
C. `x gt 1.32`
D. All real numbers

Explanation

To find the domain of the function \( f(x) = \sqrt{x^2 + 3} \), we need to determine the values of \( x \) for which this function is defined.

### Step 1: Identify the expression under the square root
The expression under the square root is \( x^2 + 3 \).

### Step 2: Analyze the expression
We know that the square of any real number \( x \) (i.e., \( x^2 \)) is always non-negative, meaning \( x^2 \geq 0 \). Therefore, \( x^2 + 3 \) will always be greater than or equal to 3, since we are adding 3 to a non-negative number.

### Step 3: Determine when the expression is non-negative
Since \( x^2 + 3 \) is always greater than or equal to 3, it is always positive. This means that the expression under the square root is always non-negative for all real values of \( x \).

### Step 4: Conclusion about the domain
Since the square root function is defined for all non-negative numbers, and since \( x^2 + 3 \) is always positive, the function \( f(x) = \sqrt{x^2 + 3} \) is defined for all real numbers.

### Final Answer
Thus, the domain of the function \( f(x) = \sqrt{x^2 + 3} \) is all real numbers, which can be expressed in interval notation as:
\[ \text{Domain} = (-\infty, \infty) \]

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