To determine the domain of the function [tex]\( f(x) = \frac{1}{\sqrt{x+3}} \)[/tex], we need to consider where the function is defined and where it makes sense mathematically.
### Step-by-Step Solution:
1. Identify restrictions due to the square root:
- The square root function, [tex]\( \sqrt{u} \)[/tex], is only defined for values [tex]\( u \geq 0 \)[/tex]. For the expression [tex]\( \sqrt{x + 3} \)[/tex] to be defined, we must have:
[tex]\[
x + 3 \geq 0
\][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
x \geq -3
\][/tex]
2. Identify restrictions due to the denominator:
- The function [tex]\( f(x) = \frac{1}{u} \)[/tex] is undefined when [tex]\( u = 0 \)[/tex] because division by zero is undefined. For our function [tex]\( f(x) = \frac{1}{\sqrt{x + 3}} \)[/tex], the denominator is [tex]\( \sqrt{x + 3} \)[/tex]. We need to ensure that:
[tex]\[
\sqrt{x + 3} \neq 0
\][/tex]
- Solving for [tex]\( x + 3 \neq 0 \)[/tex]:
[tex]\[
\sqrt{x + 3} \neq 0 \implies x + 3 \neq 0 \implies x \neq -3
\][/tex]
By combining these two conditions:
- [tex]\( x + 3 \geq 0 \)[/tex] (or [tex]\( x \geq -3 \)[/tex])
- [tex]\( x \neq -3 \)[/tex].
We find that:
- [tex]\( x \)[/tex] must be greater than [tex]\(-3\)[/tex] for the function to be defined.
### Writing the Domain:
The domain of [tex]\( f(x) = \frac{1}{\sqrt{x+3}} \)[/tex] in interval notation is:
[tex]\[
(-\infty, -3)
\][/tex]
Thus, based on these conditions, the domain of the function is:
[tex]\( x > -3 \)[/tex]
Or formally written:
[tex]\( (-\infty, -3) \)[/tex]
