To determine the intervals on which the function [tex]\( h(x) = \left|\frac{2}{\sqrt{x} - 4}\right| \)[/tex] is continuous, we need to analyze the function step-by-step.
1. Identify the function within the absolute value:
The expression inside the absolute value is [tex]\( \frac{2}{\sqrt{x} - 4} \)[/tex].
2. Determine the conditions under which this expression is defined:
The function [tex]\( \frac{2}{\sqrt{x} - 4} \)[/tex] is defined as long as its denominator is not zero.
Therefore, we need to solve for:
[tex]\[
\sqrt{x} - 4 \neq 0
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
\sqrt{x} \neq 4
\][/tex]
Squaring both sides, we get:
[tex]\[
x \neq 16
\][/tex]
4. Determine the domain of the function:
For [tex]\( \sqrt{x} \)[/tex] to be real, [tex]\( x \)[/tex] must be non-negative. Moreover, [tex]\( x \neq 16 \)[/tex].
Thus, the domain of [tex]\( \frac{2}{\sqrt{x} - 4} \)[/tex] is [tex]\( x \geq 0 \)[/tex] with the exception of [tex]\( x = 16 \)[/tex]. So, the actual domain is:
[tex]\[
x \in [0, 16) \cup (16, \infty)
\][/tex]
5. Use the continuity of the absolute value function:
Since the absolute value function [tex]\( |x| \)[/tex] is continuous for all real numbers, the continuity of [tex]\( h(x) \)[/tex] will be determined solely by the domain where [tex]\( \frac{2}{\sqrt{x} - 4} \)[/tex] is defined.
6. Combine the results:
The function [tex]\( h(x) = \left|\frac{2}{\sqrt{x} - 4}\right| \)[/tex] is continuous wherever [tex]\( \frac{2}{\sqrt{x} - 4} \)[/tex] is defined and non-zero (because the absolute value function itself is continuous). This means:
- The function is continuous for [tex]\( x \in [0, 16) \cup (16, \infty) \)[/tex].
Therefore, the function [tex]\( h(x) = \left|\frac{2}{\sqrt{x} - 4}\right| \)[/tex] is continuous on the intervals:
[tex]\[
(-\infty, 16) \cup (16, \infty)
\][/tex]
