Answer
$(-\infty,-1)\cup(0,\infty)$
or $\{x|x \lt -1$ or $x \gt 0\}$
Work Step by Step
First, we see that if we want $\displaystyle \frac{x+1}{x}$ to be defined, we must have
$x\neq 0\quad $
Next,
the argument of a logarithmic function must be greater than zero.
So, for g to be defined, we must have
$\displaystyle \frac{x+1}{x} \gt 0 \quad $
... (x+1) changes signs at -1, x changes signs at 0
$Q(x)=\displaystyle \frac{x+1}{x}$ is zero or undefined for x=-1, 0.
$\left[\begin{array}{llll}
\text{interval} & (-\infty,-1) & (-1,0) & (0,\infty)\\
\text{test point} & -2 & -0.5 & 1\\
Q(t) & \frac{-2+1}{-2} & \frac{-0.5+1}{-0.5} & \frac{1+1}{1}\\
\text{sign of }Q(t) & + & - & +
\end{array}\right]$
Domain: $(-\infty,-1)\cup(0,\infty)$
or $\{x|x \lt -1$ or $x \gt 0\}$
