Given functions \( p(x)=\frac{1}{\sqrt{x}} \) and \( h(x)=x^{2}-4 \), state the domains of the following functions using interval notation. Domain of \( \frac{p(x)}{h(x)}: \square \) Domain of \( p(h(x)): \) Question Help: Dideo \( \square \) Message instructor

To determine the domain of \( \frac{p(x)}{h(x)} \), we must find where both \( p(x) \) and \( h(x) \) are defined. The function \( p(x) = \frac{1}{\sqrt{x}} \) requires \( x > 0 \) because the square root must be positive, while the function \( h(x) = x^2 - 4 \) is defined for all real numbers but becomes zero at \( x = 2 \) and \( x = -2 \). Thus, \( h(x) \) cannot be zero, so we find that \( \frac{p(x)}{h(x)} \) is undefined at \( x = 2 \) and \( x = -2 \). Hence, the domain of \( \frac{p(x)}{h(x)} \) is \( (0, -2) \cup (-2, 2) \cup (2, \infty) \). For \( p(h(x)) \), we must first determine where \( h(x) \) is positive since \( p(x) \) is only defined for positive inputs. The function \( h(x) = x^2 - 4 \) is positive when \( x > 2 \) or \( x < -2 \). Thus, the domain of \( p(h(x)) \) is \( (-\infty, -2) \cup (2, \infty) \).