To find $f(-3)$ for the function:

$f(x) = \frac{2x^2 - 7x + 1}{3x^3 - 3}$

Substitute $x = -3$ into the function:

Step 1: Calculate the numerator

The numerator is $2x^2 - 7x + 1$.

For $x = -3$: $2(-3)^2 - 7(-3) + 1 = 2(9) + 21 + 1 = 18 + 21 + 1 = 40$

Step 2: Calculate the denominator

The denominator is $3x^3 - 3$.

For $x = -3$: $3(-3)^3 - 3 = 3(-27) - 3 = -81 - 3 = -84$

Step 3: Compute $f(-3)$

Now, divide the numerator by the denominator: $f(-3) = \frac{40}{-84} = \frac{40}{-84} = \frac{10}{-21}$

Thus, $f(-3) = -\frac{10}{21}$.

Would you like further details or explanations?

Here are some related questions:

1. How would you find the derivative of $f(x)$?
2. What happens to $f(x)$ as $x \to \infty$?
3. How would you determine where $f(x) = 0$?
4. Can $f(x)$ have vertical asymptotes? How can you find them?
5. How does changing the coefficient of $x^2$ in the numerator affect $f(x)$?

Tip: Always simplify fractions when solving rational functions to make your final answer clear.