To determine the domain of the function [tex]\(f(x) = \frac{3}{x+2} - \sqrt{x-3}\)[/tex], we need to consider the restrictions imposed by each component of the function.
1. For the term [tex]\(\frac{3}{x+2}\)[/tex]:
- This term involves a fraction, and for the fraction to be defined, the denominator cannot be zero.
- So, [tex]\(x + 2 \neq 0 \Rightarrow x \neq -2\)[/tex].
2. For the term [tex]\(\sqrt{x-3}\)[/tex]:
- This term involves a square root, and for the square root to be defined, the expression inside the square root must be non-negative.
- So, [tex]\(x - 3 \geq 0 \Rightarrow x \geq 3\)[/tex].
Combining these two conditions:
- The first condition [tex]\(x \neq -2\)[/tex] will be naturally satisfied by the second condition [tex]\(x \geq 3\)[/tex].
- Since [tex]\(x \geq 3\)[/tex] excludes any possibility of [tex]\(x\)[/tex] being less than 3, it automatically avoids [tex]\(x = -2\)[/tex] as well.
Therefore, the domain of [tex]\(f(x)\)[/tex] is all real numbers greater than or equal to 3.
Hence, the complete statement is:
The domain for [tex]\(f(x)\)[/tex] is all real numbers greater than or equal to 3.
