We all know that $0$ is excluded from the denominator of a fraction. What value of $x$, if any, would make the denominator $x^2 + 1 = 0$?

Consider the domain of each part of the subtraction problem separately.
Since $g(x)$ is the difference between $\frac{1}{x}$ and $\frac{x}{x+1}$, both excluded values must be excluded from $g(x)$.

Typically, a square root function has a domain of $x ≥ 0$, but in this case, $x$ is squared... meaning both positive or negative values will yield positive outputs. However, do not neglect to consider the $-9$.
What values of $x$, both positive and negative, will make $x^2 - 9 > 0$?

$x ≤ − 3\text{ U } x ≥ 3$

Consider the domain of the numerator and the domain of the denominator separately.
I. What is the domain of $\operatorname{log}(x - 3)$?
III. What value is excluded from the denominator of all fractions?
IV. Now combine these results to find the domain of $k(x)$.