We are given two functions, $f(x) = x - 3$ and $g(x) = \sqrt{3x - 4}$, and we need to find:

1. The function $\frac{f}{g}$
2. The function $f - g$

Additionally, we need to determine the domain of each resulting function using interval notation.

1. Finding $\frac{f}{g}$:

$\frac{f}{g}(x) = \frac{x - 3}{\sqrt{3x - 4}}$

Domain of $\frac{f}{g}$:

For the domain of $\frac{f}{g}$, we need to consider two things:

• The square root function $\sqrt{3x - 4}$ requires that $3x - 4 \geq 0$, so: $3x \geq 4 \implies x \geq \frac{4}{3}$
• Additionally, since $g(x) = \sqrt{3x - 4}$ is in the denominator, we must ensure that $g(x) \neq 0$. Therefore: $\sqrt{3x - 4} \neq 0 \implies 3x - 4 \neq 0 \implies x \neq \frac{4}{3}$ Thus, the domain of $\frac{f}{g}$ is $\left( \frac{4}{3}, \infty \right)$.

2. Finding $f - g$:

$f - g(x) = (x - 3) - \sqrt{3x - 4}$

Domain of $f - g$:

For the domain of $f - g$, we only need to ensure that the square root $\sqrt{3x - 4}$ is defined, which means: $3x - 4 \geq 0 \implies x \geq \frac{4}{3}$ So, the domain of $f - g$ is $\left[ \frac{4}{3}, \infty \right)$.

Final Results:

1. The function $\frac{f}{g}(x) = \frac{x - 3}{\sqrt{3x - 4}}$ has a domain of $\left( \frac{4}{3}, \infty \right)$.
2. The function $f - g(x) = (x - 3) - \sqrt{3x - 4}$ has a domain of $\left[ \frac{4}{3}, \infty \right)$.

Would you like more details on the steps? Here are some related questions:

1. What happens if $g(x)$ equals zero in more general cases?
2. How do you simplify rational expressions like $\frac{f}{g}$?
3. Can $f - g(x)$ ever be negative? Why or why not?
4. What are the general rules for finding domains when square roots are involved?
5. How does the difference in domain intervals between $\frac{f}{g}$ and $f - g$ affect their graphs?

Tip: Always check for restrictions like square roots and denominators when determining domains.