To determine the domain of the function [tex]\( f(x) = \frac{3}{x^2 + 1} \)[/tex], we need to examine for which values of [tex]\( x \)[/tex] the function is defined. The function [tex]\( f(x) \)[/tex] has a fraction in its expression, and a fraction is undefined when its denominator is zero. Therefore, we'll look at the denominator [tex]\( x^2 + 1 \)[/tex].
Let's analyze the denominator:
[tex]\[ x^2 + 1 \][/tex]
For the denominator to be zero, we need to solve the equation:
[tex]\[ x^2 + 1 = 0 \][/tex]
Subtract 1 from both sides:
[tex]\[ x^2 = -1 \][/tex]
However, in the real number system, there is no real number [tex]\( x \)[/tex] such that [tex]\( x^2 = -1 \)[/tex]. The square of a real number is always non-negative, meaning [tex]\( x^2 \geq 0 \)[/tex], and hence, [tex]\( x^2 + 1 \geq 1 \)[/tex] and never zero for any real number [tex]\( x \)[/tex].
Since the denominator [tex]\( x^2 + 1 \)[/tex] is never zero for any real number, the function [tex]\( f(x) \)[/tex] is defined for all real numbers.
Therefore, the domain of the function [tex]\( f(x) = \frac{3}{x^2 + 1} \)[/tex] is all real numbers.
In conclusion:
[tex]\[ \boxed{\text{all real numbers}} \][/tex]
