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

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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 $$p(h(x))$$: $$(-\infty,-2)\cup(2,\infty)$$
Domain of $$\frac{p(x)}{h(x)}$$: $$(0,2) \cup (2, \infty)$$ Domain of $$p(h(x))$$: $$(-\infty, -2) \cup (2, \infty)$$

Given functions and , state the domains of the following functions using
interval notation.
Domain of
Domain of
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