Question

How do find the domain and range of le="\frac{1}{\sqrt{x-8}}" alt="\frac{1}{\sqrt{x-8}}" align="absmiddle" class="latex-formula">?

Answer 1

$\sqrt{x-8}$ is undefined if the argument $x-8$ is negative, so you first need to require that

$x-8\ge0\implies x\ge8$

We're not done yet, though, because $\dfrac1{\sqrt{x-8}}$ still doesn't exist when $x=8$, so we remove this from the domain and we're left with $x>8$, or in interval notation, $(8,\infty)$

To find the range, consider the limits of the function as you approach either endpoint of the domain.

$\displaystyle\lim_{x\to8^+}\frac1{\sqrt{x-8}}=+\infty$
$\displaystyle\lim_{x\to\infty}\frac1{\sqrt{x-8}}=0$

Since $\dfrac1{\sqrt{x-8}}$ is positive everywhere, the range is $(0,\infty)$