To determine the domain of the function [tex]\( f(x) = \sqrt{3 - x} \)[/tex], we need to ensure that the expression inside the square root is non-negative, as the square root of a negative number is not defined in the realm of real numbers.
Here’s a detailed step-by-step process to find the domain:
1. Identify the Expression Inside the Square Root:
The function is [tex]\( f(x) = \sqrt{3 - x} \)[/tex]. The expression inside the square root is [tex]\( 3 - x \)[/tex].
2. Set Up the Inequality:
To ensure the square root is defined, the expression [tex]\( 3 - x \)[/tex] must be greater than or equal to 0. This gives us the inequality:
[tex]\[
3 - x \geq 0
\][/tex]
3. Solve the Inequality:
Solve for [tex]\( x \)[/tex] by isolating [tex]\( x \)[/tex]:
[tex]\[
3 - x \geq 0
\][/tex]
Subtract 3 from both sides:
[tex]\[
-x \geq -3
\][/tex]
Multiply both sides by -1, remembering to reverse the inequality sign:
[tex]\[
x \leq 3
\][/tex]
4. Determine the Domain:
The inequality [tex]\( x \leq 3 \)[/tex] indicates that [tex]\( x \)[/tex] can take any value less than or equal to 3. Thus, the domain includes all real numbers that are less than or equal to 3.
In set builder notation, the domain can be written as:
[tex]\[
\{ x \mid x \leq 3 \}
\][/tex]
In interval notation, the domain is:
[tex]\[
(-\infty, 3]
\][/tex]
Therefore, the domain of the function [tex]\( f(x) = \sqrt{3 - x} \)[/tex] is:
[tex]\[
(-\infty, 3] \quad \text{or} \quad \{ x \mid x \leq 3 \}
\][/tex]
