To find the domain of the rational function

$f(x) = \frac{3x(x - 1)}{2x^2 - 5x - 12}$

we need to determine the values of $x$ for which the function is defined. Since it's a rational function, the denominator must not be zero. So, we solve for the values of $x$ that make the denominator zero and exclude these values from the domain.

Step 1: Set the Denominator to Zero

The denominator of $f(x)$ is $2x^2 - 5x - 12$. Set it equal to zero and solve for $x$:

$2x^2 - 5x - 12 = 0$

Step 2: Factor the Quadratic Equation

To factor $2x^2 - 5x - 12$, we look for two numbers that multiply to $-24$ (the product of $2$ and $-12$) and add up to $-5$ (the middle coefficient).

These numbers are $-8$ and $3$, so we can rewrite the expression as:

$2x^2 - 8x + 3x - 12 = 0$

Now, factor by grouping:

$2x(x - 4) + 3(x - 4) = 0$

$(2x + 3)(x - 4) = 0$

Step 3: Solve for $x$

Now, set each factor to zero:

1. $2x + 3 = 0$ gives $x = -\frac{3}{2}$
2. $x - 4 = 0$ gives $x = 4$

Step 4: Determine the Domain

Since $f(x)$ is undefined at $x = -\frac{3}{2}$ and $x = 4$, we exclude these values from the domain.

Thus, the domain of $f(x)$ is all real numbers except $x = -\frac{3}{2}$ and $x = 4$. In interval notation, this is:

$\text{Domain of } f(x) = (-\infty, -\frac{3}{2}) \cup \left(-\frac{3}{2}, 4\right) \cup (4, \infty)$

Would you like more details on any of these steps?

Here are some related questions:

1. How do you find the domain of a function with both square roots and rational expressions?
2. What happens to the graph of a function at points excluded from its domain?
3. Can factoring always be used to find where a quadratic expression is zero?
4. How do you find the range of a rational function?
5. What are vertical asymptotes and how are they related to the domain?

Tip: Always set the denominator of a rational function equal to zero to identify points excluded from its domain.