To determine the domain of the function [tex]\( r = \sqrt{\frac{A}{\pi}} \)[/tex], we need to identify the set of all possible values for [tex]\( A \)[/tex] that make the function [tex]\( r \)[/tex] well-defined and mathematically valid.
1. Understanding the given function:
- The function [tex]\( r = \sqrt{\frac{A}{\pi}} \)[/tex] expresses the radius [tex]\( r \)[/tex] of a circle in terms of its area [tex]\( A \)[/tex].
- This formula comes from rearranging the area formula [tex]\( A = \pi r^2 \)[/tex].
2. Analyzing the expression inside the square root:
- The expression [tex]\(\frac{A}{\pi}\)[/tex] involves division by [tex]\(\pi\)[/tex], a positive constant.
3. Domain considerations for the square root:
- The square root function [tex]\(\sqrt{x}\)[/tex] is only defined for non-negative values of [tex]\( x \)[/tex]. Therefore, the expression inside the square root, [tex]\(\frac{A}{\pi}\)[/tex], must be non-negative.
- Given that [tex]\(\pi\)[/tex] is a positive constant, the fraction [tex]\(\frac{A}{\pi}\)[/tex] is non-negative if and only if [tex]\( A \)[/tex] is non-negative.
4. Conclusion:
- Since the area [tex]\( A \)[/tex] of a circle must be a non-negative number (as negative area does not make sense in the physical world), we conclude that [tex]\( A \)[/tex] must be greater than or equal to 0.
Thus, the domain of the function [tex]\( r = \sqrt{\frac{A}{\pi}} \)[/tex] is all non-negative real numbers. Hence, the domain is:
[tex]\[ [0, \infty) \][/tex]
This means that [tex]\( A \)[/tex] can take any value from 0 to positive infinity, inclusive of 0.
