What is the domain of (f(x)=\frac{2x-1}{\sqrt{9-x^2}})?

Correct answer: ((-3,3))

For a square root to be real, its radicand must be nonnegative. Here that condition is 9-x^2 greater than or equal to 0. However, the square root is in the denominator, so it cannot itself equal zero. Therefore the stronger condition is 9-x^2>0. Solving it gives x^2<9, hence -3<x<3, which is option A.

The endpoints -3 and 3 are excluded because at either endpoint the denominator becomes \\(\sqrt{0}=0\\, making the fraction undefined. Values between the endpoints make the radicand positive and are allowed. Values outside the interval make the radicand negative, so the square root is not real. Thus neither the closed interval [-3,3] nor all real numbers is correct; the domain is (-3,3).