Question

given functions $m(x)=\frac{1}{sqrt{x}}$ and $p(x)=x^{2}-4$, state the domains of the following functions using interval notation.
domain of $\frac{m(x)}{p(x)}$ :
domain of $m(p(x))$ :
domain of $p(m(x))$ :
question help: video

Explanation:

Step1: Analyze domain of $m(x)$

For $m(x)=\frac{1}{\sqrt{x}}$, $x>0$ since we can't have negative values under square - root in the denominator. So domain of $m(x)$ is $(0,+\infty)$.

Step2: Analyze domain of $p(x)$

For $p(x)=x^{2}-4$, $x\in(-\infty,+\infty)$ as it is a polynomial function.

Step3: Find domain of $\frac{m(x)}{p(x)}$

We need $x > 0$ (from $m(x)$) and $p(x)
eq0$. Solving $p(x)=x^{2}-4=(x + 2)(x - 2)=0$, we get $x=-2$ or $x = 2$. Since $x>0$, we exclude $x = 2$. So the domain is $(0,2)\cup(2,+\infty)$.

Step4: Find domain of $m(p(x))$

We need $p(x)>0$. So $x^{2}-4>0$, which gives $(x + 2)(x - 2)>0$. The solutions of the inequality are $x<-2$ or $x>2$. So the domain is $(-\infty,-2)\cup(2,+\infty)$.

Step5: Find domain of $p(m(x))$

We need $x>0$ since the input of $m(x)$ must satisfy the domain of $m(x)$. So the domain is $(0,+\infty)$.

Answer:

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