Question

which of the following is the domain of the function $f(x) = \frac{x^2 - 4}{4x^3 - 7x^2 - 2x}$? (1 point) $\bigcirc\\ (-infty, -2) \cup \left(-2, \frac{1}{4}\
ight) \cup \left(\frac{1}{4}, infty\
ight)$ $\bigcirc\\ \left(-infty, \frac{1}{4}\
ight) \cup \left(\frac{1}{4}, 0\
ight) \cup (0, infty)$ $\bigcirc\\ \left(-infty, -\frac{1}{4}\
ight) \cup \left(-\frac{1}{4}, 0\
ight) \cup (0, 2) \cup (2, infty)$ $\bigcirc\\ \left(-infty, -\frac{1}{4}\
ight) \cup \left(-\frac{1}{4}, 0\
ight) \cup (0, infty)$

Explanation:

Step1: Identify undefined condition

A rational function is undefined when its denominator equals 0. So solve:
$$4x^3 -7x^2 -2x = 0$$

Step2: Factor the denominator

Factor out $x$ first, then factor the quadratic:
$$x(4x^2 -7x -2) = 0$$
$$x(4x+1)(x-2) = 0$$

Step3: Solve for denominator roots

Set each factor equal to 0:
$x=0$, $4x+1=0 \implies x=-\frac{1}{4}$, $x-2=0 \implies x=2$

Step4: Exclude undefined values

The domain includes all real numbers except $x=-\frac{1}{4}$, $x=0$, and $x=2$, which corresponds to the interval:
$$(-\infty,-\frac{1}{4}) \cup (-\frac{1}{4},0) \cup (0,2) \cup (2,\infty)$$

Answer:

$\boldsymbol{(-\infty,-\frac{1}{4}) \cup (-\frac{1}{4},0) \cup (0,2) \cup (2,\infty)}$