To determine the range of the function [tex]\((u \cdot v)(x)\)[/tex], where [tex]\(u(x) = -2x^2 + 3\)[/tex] and [tex]\(v(x) = \frac{1}{x}\)[/tex], we need to examine the behavior of their product [tex]\((u \cdot v)(x)\)[/tex].
First, let's denote the product function as [tex]\(f(x)\)[/tex]:
[tex]\[
f(x) = u(x) \cdot v(x) = (-2x^2 + 3) \cdot \frac{1}{x}
\][/tex]
Simplifying [tex]\(f(x)\)[/tex], we get:
[tex]\[
f(x) = -2x + \frac{3}{x}
\][/tex]
To understand the range of [tex]\(f(x)\)[/tex], let's analyze the expression [tex]\(-2x + \frac{3}{x}\)[/tex].
The behavior of the function as [tex]\(x\)[/tex] varies:
1. As [tex]\(x\)[/tex] approaches 0 from the positive side ([tex]\(x \to 0^+\)[/tex]):
[tex]\[
-2x \to 0 \quad \text{and} \quad \frac{3}{x} \to +\infty
\][/tex]
Therefore, [tex]\(f(x) \to +\infty\)[/tex].
2. As [tex]\(x\)[/tex] approaches 0 from the negative side ([tex]\(x \to 0^-\)[/tex]):
[tex]\[
-2x \to 0 \quad \text{and} \quad \frac{3}{x} \to -\infty
\][/tex]
Therefore, [tex]\(f(x) \to -\infty\)[/tex].
3. As [tex]\(x\)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[
-2x \to -\infty \quad \text{and} \quad \frac{3}{x} \to 0
\][/tex]
Therefore, [tex]\(f(x) \to -\infty\)[/tex].
4. As [tex]\(x\)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[
-2x \to +\infty \quad \text{and} \quad \frac{3}{x} \to 0
\][/tex]
Therefore, [tex]\(f(x) \to +\infty\)[/tex].
Given this analysis, we can see that as [tex]\(x\)[/tex] varies over the real numbers (except zero), the function [tex]\(f(x) = -2x + \frac{3}{x}\)[/tex] takes on all real values.
Thus, the range of the function [tex]\((u \cdot v)(x)\)[/tex] is:
[tex]\[
(-\infty, +\infty)
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{(-\infty, +\infty)}
\][/tex]
