Question

find $(p \circ m)(x)$ and the domain of $(p \circ m)(x)$ for the functions.
$p(x) = \dfrac{1}{\sqrt{x}}$ and $m(x) = x^2 - 4$

select one:
\\( \bigcirc \\) a. $\\ (p \circ m)(x) = \dfrac{1}{\sqrt{x^2 - 4}}$ and domain:
\\( \quad \quad (-\infty, -2 \cup (2, \infty) \\)
\\( \bigcirc \\) b. $\\ (p \circ m)(x) = \dfrac{1}{\sqrt{x^2 - 4}}$ and domain:
\\( \quad \quad (-\infty, -2 \cup 2, \infty) \\)
\\( \bigcirc \\) c. $\\ (p \circ m)(x) = \dfrac{1}{\sqrt{x^2 - 4}}$ and domain:
\\( \quad \quad (-\infty, -2) \cup 2, \infty) \\)
\\( \bigcirc \\) d. $\\ (p \circ m)(x) = \dfrac{1}{\sqrt{x^2 - 4}}$ and domain:
\\( \quad \quad (-\infty, -2) \cup (2, \infty) \\)

Explanation:

Step1: Compute composite function

Substitute $m(x)$ into $p(x)$:
$$(p \circ m)(x) = p(m(x)) = \frac{1}{\sqrt{m(x)}} = \frac{1}{\sqrt{x^2 - 4}}$$

Step2: Find domain constraints

For $\sqrt{x^2 - 4}$:

1. Radicand must be positive (denominator ≠ 0, square root requires non-negative, but denominator can't be 0):

$$x^2 - 4 > 0$$

1. Solve the inequality:

$$x^2 > 4 \implies |x| > 2$$
This simplifies to $x < -2$ or $x > 2$, written as $(-\infty, -2) \cup (2, \infty)$

Answer:

D. $(p \circ m)(x) = \frac{1}{\sqrt{x^2-4}}$ and Domain: $(-\infty,-2)\cup(2,\infty)$