Given functions \( g(x)=\frac{1}{\sqrt{x}} \) and \( p(x)=x^{2}-4 \), state the domains of the following functions using interval notation. Domain of \( \frac{g(x)}{p(x)}:(0,2) \cup(2, \infty) \) Domain of \( g(p(x)): \)

To find the domain of \( g(p(x)) \), we first need to determine the range of \( p(x) \). The function \( p(x) = x^2 - 4 \) is a quadratic function that opens upwards, with its vertex at \( (0, -4) \). The minimum value of \( p(x) \) is -4, which occurs when \( x = 0 \). Thus, \( p(x) \) will take values from \(-4\) to \(\infty\). For the function \( g(x) = \frac{1}{\sqrt{x}} \), we require \( x > 0 \) because the square root function is defined only for non-negative numbers, and we cannot take the square root of 0 when it is in the denominator. Now we need \( p(x) > 0 \) for \( g(p(x)) \) to be defined. Solving the inequality \( p(x) > 0 \): \[ x^2 - 4 > 0 \] This factors as: \[ (x - 2)(x + 2) > 0 \] The solutions to this inequality are found by testing the intervals determined by the roots \( x = -2 \) and \( x = 2 \). The function is positive for \( x < -2 \) and \( x > 2 \). Since we also have the restriction that \( p(x) \) must be greater than 0, we combine the intervals where \( p(x) > 0 \) with \( x > 0 \). Thus, the domain of \( g(p(x)) \) is \( (2, \infty) \) in interval notation. So to summarize: Domain of \( g(p(x)): (2, \infty) \)