To determine the domain of the rational function [tex]\( f(x) = \frac{3x}{2x^3 - x^2 - 15x} \)[/tex], we need to identify the values of [tex]\( x \)[/tex] that make the denominator equal to zero, because these values are not part of the domain. The denominator is:
[tex]\[ 2x^3 - x^2 - 15x \][/tex]
We need to solve the equation [tex]\( 2x^3 - x^2 - 15x = 0 \)[/tex] for [tex]\( x \)[/tex].
First, factor out the common term [tex]\( x \)[/tex] from the denominator:
[tex]\[ 2x^3 - x^2 - 15x = x(2x^2 - x - 15) \][/tex]
We have one solution from the factored out [tex]\( x \)[/tex]:
[tex]\[ x = 0 \][/tex]
Now we need to find the roots of the quadratic [tex]\( 2x^2 - x - 15 \)[/tex]. To do this, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 2 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -15 \)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-15)}}{2 \cdot 2} \][/tex]
This simplifies to:
[tex]\[ x = \frac{1 \pm \sqrt{1 + 120}}{4} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{121}}{4} \][/tex]
[tex]\[ x = \frac{1 \pm 11}{4} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{1 + 11}{4} = \frac{12}{4} = 3 \][/tex]
[tex]\[ x = \frac{1 - 11}{4} = \frac{-10}{4} = -\frac{5}{2} \][/tex]
Thus, the values of [tex]\( x \)[/tex] that make the denominator zero are [tex]\( x = 0 \)[/tex], [tex]\( x = 3 \)[/tex], and [tex]\( x = -\frac{5}{2} \)[/tex].
Therefore, the domain of the rational function is all real numbers except these values. In set notation, the domain is:
[tex]\[ \{ x \in \mathbb{R} \mid x \neq 0, 3, -\frac{5}{2}\} \][/tex]
Comparing this with the given options, the correct answer is:
[tex]\[ \left\{ x \in \mathbb{R} \mid x \neq 0, 3, -\frac{5}{2} \right\} \][/tex]
