4. What is the (largest) domain of $f(x)=\dfrac{1}{x-1}$ (in the real numbers)? Solution: There are no even roots so we will focus on the denominator. We must identify the $x$'s which make the denominator zero and then remove the mfrom consideration: set the denominator equal to zero to arrive at the equation
$$x-1=0.$$
This equation has solution $x=1$. Therefore the domain is "all readl numbers except for $1$", or written in other ways: $(-\infty,1) \cup (1,\infty)$, or $-\infty \lt x \lt 1$ and $1 \lt x \lt \infty$, or $\mathbb{R} \setminus \{1\}$.
5. What is the (largest) domain of $f(x) = \dfrac{\sqrt{x-1}}{3x+2}$ (in the real numbers)? Solution: We must consider both the root and the denominator. First we find the $x$'s that make the denominator equation to zero by setting the denominator equal to zero and solving:
$$3x+2=0,$$
so
$$3x=-2,$$
so
$$x=-\dfrac{2}{3}.$$
We will have to remove $-\dfrac{2}{3}$ from the domain. Now we must ensure that what appears under the root is non-negative (i.e. $\geq 0$):
$$x-1 \geq 0,$$
or by adding $1$,
$$x \geq 1.$$
Therefore the domain is "all real numbers greater than or equal to $1$ but not $-\dfrac{2}{3}$." note: the statement "all real numbers greater than or equal to $1$" already excludes the point $-\dfrac{2}{3}$, so it would be ok to say just that in this circumstance