To determine the domain of the function [tex]\( f(x) = \frac{1}{\sqrt{4x - 12}} \)[/tex], we need to ensure that the expression under the square root is positive because the square root of a non-positive number is not defined in the real numbers, and division by zero is undefined.
The expression under the square root, [tex]\( 4x - 12 \)[/tex], must be strictly greater than zero. We set up the following inequality:
[tex]\[ 4x - 12 > 0 \][/tex]
Solving this inequality for [tex]\( x \)[/tex]:
1. Add 12 to both sides:
[tex]\[ 4x - 12 + 12 > 0 + 12 \][/tex]
[tex]\[ 4x > 12 \][/tex]
2. Divide both sides by 4:
[tex]\[ \frac{4x}{4} > \frac{12}{4} \][/tex]
[tex]\[ x > 3 \][/tex]
So, [tex]\( x \)[/tex] must be greater than 3 for the function [tex]\( f(x) = \frac{1}{\sqrt{4x - 12}} \)[/tex] to be defined.
Therefore, the domain of the function in interval notation is:
[tex]\[ (3, \infty) \][/tex]
This indicates that [tex]\( x \)[/tex] can be any real number greater than 3.
