Answer :
Given the functions [tex]\( u(x) = -2x^2 + 3 \)[/tex] and [tex]\( v(x) = \frac{1}{x} \)[/tex], we need to determine the range of the product function [tex]\( (u \cdot v)(x) \)[/tex].
First, let's define the product function [tex]\( (u \cdot v)(x) \)[/tex]:
[tex]\[ (u \cdot v)(x) = u(x) \cdot v(x) = \left(-2x^2 + 3\right) \cdot \left(\frac{1}{x}\right) = \frac{-2x^2 + 3}{x} \][/tex]
To simplify:
[tex]\[ (u \cdot v)(x) = \frac{-2x^2 + 3}{x} = -2x + \frac{3}{x} \][/tex]
Next, we need to analyze the range of the function [tex]\( -2x + \frac{3}{x} \)[/tex].
We can examine the behavior of this function as [tex]\( x \)[/tex] approaches different values:
1. As [tex]\( x \)[/tex] approaches 0 from the positive side ([tex]\( x \to 0^+ \)[/tex]):
[tex]\[ -2x \to 0 \text{ and } \frac{3}{x} \to +\infty \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to +\infty \)[/tex].
2. As [tex]\( x \)[/tex] approaches 0 from the negative side ([tex]\( x \to 0^- \)[/tex]):
[tex]\[ -2x \to 0 \text{ and } \frac{3}{x} \to -\infty \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to -\infty \)[/tex].
3. As [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] ([tex]\( x \to \infty \)[/tex]):
[tex]\[ -2x \to -\infty \text{ and } \frac{3}{x} \to 0 \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to -\infty \)[/tex].
4. As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] ([tex]\( x \to -\infty \)[/tex]):
[tex]\[ -2x \to +\infty \text{ and } \frac{3}{x} \to 0 \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to +\infty \)[/tex].
From these observations, we see that the function [tex]\( -2x + \frac{3}{x} \)[/tex] can take on any real value. It can approach [tex]\( +\infty \)[/tex] as [tex]\( x \)[/tex] approaches 0 from the positive side or [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches 0 from the negative side. It can also go to [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and to [tex]\( +\infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex].
Thus, the range of [tex]\( (u \cdot v)(x) \)[/tex] is:
[tex]\[ (-\infty, +\infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{(-\infty, +\infty)} \][/tex]
First, let's define the product function [tex]\( (u \cdot v)(x) \)[/tex]:
[tex]\[ (u \cdot v)(x) = u(x) \cdot v(x) = \left(-2x^2 + 3\right) \cdot \left(\frac{1}{x}\right) = \frac{-2x^2 + 3}{x} \][/tex]
To simplify:
[tex]\[ (u \cdot v)(x) = \frac{-2x^2 + 3}{x} = -2x + \frac{3}{x} \][/tex]
Next, we need to analyze the range of the function [tex]\( -2x + \frac{3}{x} \)[/tex].
We can examine the behavior of this function as [tex]\( x \)[/tex] approaches different values:
1. As [tex]\( x \)[/tex] approaches 0 from the positive side ([tex]\( x \to 0^+ \)[/tex]):
[tex]\[ -2x \to 0 \text{ and } \frac{3}{x} \to +\infty \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to +\infty \)[/tex].
2. As [tex]\( x \)[/tex] approaches 0 from the negative side ([tex]\( x \to 0^- \)[/tex]):
[tex]\[ -2x \to 0 \text{ and } \frac{3}{x} \to -\infty \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to -\infty \)[/tex].
3. As [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] ([tex]\( x \to \infty \)[/tex]):
[tex]\[ -2x \to -\infty \text{ and } \frac{3}{x} \to 0 \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to -\infty \)[/tex].
4. As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] ([tex]\( x \to -\infty \)[/tex]):
[tex]\[ -2x \to +\infty \text{ and } \frac{3}{x} \to 0 \][/tex]
Therefore, [tex]\( -2x + \frac{3}{x} \to +\infty \)[/tex].
From these observations, we see that the function [tex]\( -2x + \frac{3}{x} \)[/tex] can take on any real value. It can approach [tex]\( +\infty \)[/tex] as [tex]\( x \)[/tex] approaches 0 from the positive side or [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches 0 from the negative side. It can also go to [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] and to [tex]\( +\infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex].
Thus, the range of [tex]\( (u \cdot v)(x) \)[/tex] is:
[tex]\[ (-\infty, +\infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{(-\infty, +\infty)} \][/tex]
