Let's analyze the given functions one by one to determine their domains:
1. Function [tex]\( f(x) = x^2 + 3 \)[/tex]
- This function is a polynomial function. Polynomial functions are defined for all real numbers because you can square any real number and add 3 to it without any restrictions.
- Domain of [tex]\( f \)[/tex]: All real numbers [tex]\(( -\infty, \infty )\)[/tex]
2. Function [tex]\( g(x) = \frac{1}{x-1} \)[/tex]
- This function is a rational function. Rational functions are defined everywhere their denominators are non-zero. In this case, the denominator is [tex]\( x - 1 \)[/tex]. This expression is zero when [tex]\( x = 1 \)[/tex], making the function undefined at [tex]\( x = 1 \)[/tex].
- Domain of [tex]\( g \)[/tex]: All real numbers except [tex]\( x = 1 \)[/tex]
3. Function [tex]\( h(x) = \sqrt{x-2} \)[/tex]
- This function is a square root function. Square root functions are defined for non-negative arguments because you cannot take the square root of a negative number. Here, the argument under the square root is [tex]\( x - 2 \)[/tex]. This expression is non-negative when [tex]\( x \geq 2 \)[/tex].
- Domain of [tex]\( h \)[/tex]: [tex]\( x \geq 2 \)[/tex] or [tex]\([2, \infty)\)[/tex]
Now, we need to determine which functions have a domain of all real numbers:
- [tex]\( f(x) \)[/tex] has a domain of all real numbers.
- [tex]\( g(x) \)[/tex] does not have a domain of all real numbers because it is undefined at [tex]\( x = 1 \)[/tex].
- [tex]\( h(x) \)[/tex] does not have a domain of all real numbers because it is only defined for [tex]\( x \geq 2 \)[/tex].
Only the function [tex]\( f(x) \)[/tex] has a domain of all real numbers.
Therefore, the correct answer is:
- A. function [tex]\( f \)[/tex] only
