Given functions \( m(x)=\frac{1}{\sqrt{x}} \) and \( g(x)=x^{2}-36 \), state the domains of the following functions using interval notation. Domain of \( \frac{m(x)}{g(x)}: \square \) Domain of \( m(g(x)): \square \) Domain of \( g(m(x)): \)

To determine the domains of the functions, we need to consider the restrictions for each function: 1. For \( \frac{m(x)}{g(x)} \), the function is undefined when the denominator \( g(x) = x^2 - 36 \) is equal to zero. This occurs when \( x^2 = 36 \), or \( x = 6 \) and \( x = -6 \). Additionally, since \( m(x) = \frac{1}{\sqrt{x}} \) requires \( x > 0 \), the domain of \( \frac{m(x)}{g(x)} \) excludes \( x = 6 \) and \( x = -6 \) and must also be greater than 0. So, the domain is \( (0, 6) \cup (6, \infty) \). 2. For \( m(g(x)) \), we need \( g(x) \) to be positive because \( m(x) \) requires the argument \( g(x) \) to be greater than 0. The function \( g(x) = x^2 - 36 \) is greater than 0 when \( x < -6 \) or \( x > 6 \). Therefore, the domain of \( m(g(x)) \) is \( (-\infty, -6) \cup (6, \infty) \). 3. For \( g(m(x)) \), we first find \( m(x) = \frac{1}{\sqrt{x}} \), which is only defined for \( x > 0 \). Thus, we take any \( x > 0 \) and substitute it into \( g(m(x)) \). Since there are no further restrictions coming from \( g(x) \), we can plug in values for \( m(x) \) and see that it outputs values from 0 onward, as \( m(x) \) approaches infinity as \( x \) approaches 0. Therefore, the domain for \( g(m(x)) \) confirms \( x > 0 \), or \( (0, \infty) \). So, the final answers are: Domain of \( \frac{m(x)}{g(x)}: (0, 6) \cup (6, \infty) \) Domain of \( m(g(x)): (-\infty, -6) \cup (6, \infty) \) Domain of \( g(m(x)): (0, \infty) \)