The problem asks us to evaluate $\left( \frac{f}{g} \right)(2)$, where:

• $f(x) = x^3 + 2x - 3$
• $g(x) = 3x - 2$

Steps:

1. First, we need to find $f(2)$ and $g(2)$.

   For $f(x) = x^3 + 2x - 3$, substitute $x = 2$: $f(2) = 2^3 + 2(2) - 3 = 8 + 4 - 3 = 9$

   For $g(x) = 3x - 2$, substitute $x = 2$: $g(2) = 3(2) - 2 = 6 - 2 = 4$

2. Now, compute $\left( \frac{f}{g} \right)(2)$, which is $\frac{f(2)}{g(2)}$: $\left( \frac{f}{g} \right)(2) = \frac{9}{4}$

Thus, the answer is $\frac{9}{4}$.

Would you like more details or have any questions?

Related Questions:

1. How do you find the domain of $\frac{f(x)}{g(x)}$?
2. What happens if $g(x) = 0$ at some value of $x$?
3. Can we simplify the fraction further if it were different?
4. How do you find the derivative of $\frac{f(x)}{g(x)}$?
5. How does $\frac{f(x)}{g(x)}$ behave as $x$ approaches infinity?

Tip: Always ensure the denominator $g(x)$ is non-zero before dividing!