To determine the domain of the function [tex]\( f(x) = \sqrt{x^2 + 2} \)[/tex], we need to find all the values of [tex]\( x \)[/tex] for which the expression inside the square root is non-negative. A square root function is defined only for non-negative values under the radical.
Let's consider the expression inside the square root, [tex]\( x^2 + 2 \)[/tex].
1. Expression Analysis: We need [tex]\( x^2 + 2 \geq 0 \)[/tex].
2. Evaluate the Expression: Since [tex]\( x^2 \)[/tex] represents the square of [tex]\( x \)[/tex], it is always non-negative for all real numbers [tex]\( x \)[/tex]. Specifically, [tex]\( x^2 \geq 0 \)[/tex] for all [tex]\( x \in \mathbb{R} \)[/tex].
3. Adding 2: When 2 is added to any non-negative number, the result is always positive. Therefore, [tex]\( x^2 + 2 \geq 2 > 0 \)[/tex]. This means that the expression [tex]\( x^2 + 2 \)[/tex] is always positive regardless of the value of [tex]\( x \)[/tex].
4. Conclusion: Since [tex]\( x^2 + 2 \)[/tex] is always positive for all real numbers, the square root function [tex]\( \sqrt{x^2 + 2} \)[/tex] is defined for all real numbers [tex]\( x \)[/tex].
Thus, the domain of the function [tex]\( f(x) = \sqrt{x^2 + 2} \)[/tex] is all real numbers. In interval notation, this is represented as:
[tex]\[ \text{Domain} = (-\infty, \infty) \][/tex]
