Answer :
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], follow these steps:
1. Define the Composite Function:
[tex]\[ (u \cdot v)(x) = \left(-2x^2 + 3\right) \cdot \frac{1}{x} = \frac{-2x^2 + 3}{x} \][/tex]
2. Simplify the Composite Function:
[tex]\[ (u \cdot v)(x) = \frac{-2x^2 + 3}{x} = -2x + \frac{3}{x} \][/tex]
3. Find the Critical Points:
To find the critical points, compute the derivative of [tex]\((u \cdot v)(x)\)[/tex] and set it to zero:
[tex]\[ f(x) = -2x + \frac{3}{x} \][/tex]
[tex]\[ f'(x) = -2 - \frac{3}{x^2} \][/tex]
Set the derivative equal to zero:
[tex]\[ -2 - \frac{3}{x^2} = 0 \quad \Rightarrow\quad -2 = \frac{3}{x^2} \quad \Rightarrow\quad x^2 = -\frac{3}{2} \][/tex]
Since [tex]\(x^2\)[/tex] cannot be negative, there are no real solutions to this equation. Therefore, there are no critical points in the real number domain.
4. Evaluate the Function as [tex]\(x\)[/tex] Approaches Infinity:
Examine the behavior of the function as [tex]\(x \to \infty\)[/tex] and [tex]\(x \to -\infty\)[/tex]:
[tex]\[ \lim_{x \to \infty} \left(-2x + \frac{3}{x}\right) = -\infty \][/tex]
[tex]\[ \lim_{x \to -\infty} \left(-2x + \frac{3}{x}\right) = \infty \][/tex]
5. Evaluate the Function as [tex]\(x\)[/tex] Approaches Zero:
Examine the behavior of the function as [tex]\(x \to 0\)[/tex] from both sides:
[tex]\[ \lim_{x \to 0^+} \left(-2x + \frac{3}{x}\right) = \infty \][/tex]
[tex]\[ \lim_{x \to 0^-} \left(-2x + \frac{3}{x}\right) = -\infty \][/tex]
From these observations, it is clear that the function [tex]\((u \cdot v)(x) = -2x + \frac{3}{x}\)[/tex] takes on all real values as [tex]\(x\)[/tex] ranges over all real numbers except zero. Thus, the range of [tex]\((u \cdot v)(x)\)[/tex] is [tex]\(\boxed{(-\infty, +\infty)}\)[/tex].
1. Define the Composite Function:
[tex]\[ (u \cdot v)(x) = \left(-2x^2 + 3\right) \cdot \frac{1}{x} = \frac{-2x^2 + 3}{x} \][/tex]
2. Simplify the Composite Function:
[tex]\[ (u \cdot v)(x) = \frac{-2x^2 + 3}{x} = -2x + \frac{3}{x} \][/tex]
3. Find the Critical Points:
To find the critical points, compute the derivative of [tex]\((u \cdot v)(x)\)[/tex] and set it to zero:
[tex]\[ f(x) = -2x + \frac{3}{x} \][/tex]
[tex]\[ f'(x) = -2 - \frac{3}{x^2} \][/tex]
Set the derivative equal to zero:
[tex]\[ -2 - \frac{3}{x^2} = 0 \quad \Rightarrow\quad -2 = \frac{3}{x^2} \quad \Rightarrow\quad x^2 = -\frac{3}{2} \][/tex]
Since [tex]\(x^2\)[/tex] cannot be negative, there are no real solutions to this equation. Therefore, there are no critical points in the real number domain.
4. Evaluate the Function as [tex]\(x\)[/tex] Approaches Infinity:
Examine the behavior of the function as [tex]\(x \to \infty\)[/tex] and [tex]\(x \to -\infty\)[/tex]:
[tex]\[ \lim_{x \to \infty} \left(-2x + \frac{3}{x}\right) = -\infty \][/tex]
[tex]\[ \lim_{x \to -\infty} \left(-2x + \frac{3}{x}\right) = \infty \][/tex]
5. Evaluate the Function as [tex]\(x\)[/tex] Approaches Zero:
Examine the behavior of the function as [tex]\(x \to 0\)[/tex] from both sides:
[tex]\[ \lim_{x \to 0^+} \left(-2x + \frac{3}{x}\right) = \infty \][/tex]
[tex]\[ \lim_{x \to 0^-} \left(-2x + \frac{3}{x}\right) = -\infty \][/tex]
From these observations, it is clear that the function [tex]\((u \cdot v)(x) = -2x + \frac{3}{x}\)[/tex] takes on all real values as [tex]\(x\)[/tex] ranges over all real numbers except zero. Thus, the range of [tex]\((u \cdot v)(x)\)[/tex] is [tex]\(\boxed{(-\infty, +\infty)}\)[/tex].
