Answer
a.$\displaystyle \quad q(x)=\frac{\sqrt{10-x}}{x-1}$
b.$\quad [0,1)\cup(1,10]$
c.$\quad$undefined
Work Step by Step
$a.$
$q=\displaystyle \frac{f}{g}$ is the function specified by $q(x)=\displaystyle \frac{f(x)}{g(x)}$
$q(x)=\displaystyle \frac{\sqrt{10-x}}{x-1}$
$b.$
See Note on Domains p.70-71
Domain of $f/g$:
All real numbers $x$ simultaneously in the domains of $f$ and $g$ such that $g(x)\neq 0$
Given the domains of $v$ and $g$,
the domain of $q$ is the set
$[0,10]$ for which $x-1\neq 0$,
$[0,10]$ for which $x\neq 1$.
That is, $0 \leq x <1 $ or $1< x \leq 10$,
or written as
$[0,1)\cup(1,10]$
$\mathrm{c}.$
$q(1)$ is not defined, as 1 is not in the domain.,
(yields zero in the denominator)
