To find the domain of the function
\[ f(x) = \frac{x^2 - 4}{4x^3 - 7x^2 - 2x} \]
we need to identify the values of \(x\) that would make the denominator zero, as these are the values that are not included in the domain.
First, let's set the denominator equal to zero and solve for \(x\):
\[ 4x^3 - 7x^2 - 2x = 0 \]
We can factor out \(x\):
\[ x(4x^2 - 7x - 2) = 0 \]
This gives us one solution: \(x = 0\).
Next, we need to solve the quadratic equation \(4x^2 - 7x - 2 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 4\), \(b = -7\), and \(c = -2\):
\[ x = \frac{7 \pm \sqrt{(-7)^2 - 4(4)(-2)}}{2(4)} = \frac{7 \pm \sqrt{49 + 32}}{8} = \frac{7 \pm \sqrt{81}}{8} = \frac{7 \pm 9}{8} \]
This gives us two potential solutions:
\[ x = \frac{16}{8} = 2 \quad \text{and} \quad x = \frac{-2}{8} = -\frac{1}{4} \]
Now we have three values where the denominator is zero:
- \( x = 0 \)
- \( x = 2 \)
- \( x = -\frac{1}{4} \)
Thus, the function is undefined at \(x = 0\), \(x = 2\), and \(x = -\frac{1}{4}\).
The domain of the function can be expressed in interval notation, excluding the points where the function is undefined:
\[ (-\infty, -\frac{1}{4}) \cup (-\frac{1}{4}, 0) \cup (0, 2) \cup (2, \infty) \]
Thus, the relevant option from your responses that corresponds to the domain is:
\[ \text{left parenthesis negative infinity comma negative one fourth right parenthesis Union left parenthesis negative one fourth comma 0 right parenthesis Union left parenthesis 0 comma 2 right parenthesis Union left parenthesis 2 comma infinity right parenthesis} \]