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

1. For the function \(\frac{p(x)}{h(x)}\): - We have \(p(x)=\frac{1}{\sqrt{x}}\), so the domain of \(p(x)\) requires the expression under the square root to be positive: \[ x>0. \] - Also, for \[ h(x)=x^2-4, \] we require the denominator \(h(x)\) to be nonzero: \[ x^2-4\neq 0 \quad \Longrightarrow \quad x^2\neq 4 \quad \Longrightarrow \quad x\neq 2 \text{ and } x\neq -2. \] Since \(x>0\), the restriction \(x\neq -2\) is automatically satisfied. - Combining these, the domain of \(\frac{p(x)}{h(x)}\) is: \[ (0,2)\cup(2,\infty). \] 2. For the function \(p(h(x))\): - The expression is \[ p(h(x))=\frac{1}{\sqrt{h(x)}}=\frac{1}{\sqrt{x^2-4}}. \] - The square root requires: \[ x^2-4\ge 0 \quad \Longrightarrow \quad x^2\ge 4 \quad \Longrightarrow \quad x\le -2 \text{ or } x\ge 2. \] - Moreover, the denominator of the composed function must not be zero: \[ \sqrt{x^2-4}\neq 0 \quad \Longrightarrow \quad x^2-4\neq 0 \quad \Longrightarrow \quad x\neq -2 \text{ and } x\neq 2. \] - Thus, the domain of \(p(h(x))\) is: \[ (-\infty,-2)\cup(2,\infty). \] **Final Answers:** - Domain of \(\frac{p(x)}{h(x)}\): \((0,2)\cup(2,\infty)\) - Domain of \(p(h(x))\): \((-\infty,-2)\cup(2,\infty)\)