Answer :
To determine the range of the composite function [tex]\((u \circ v)(x)\)[/tex], where [tex]\(u(x) = -2x^2 + 3\)[/tex] and [tex]\(v(x) = \frac{1}{x}\)[/tex], we need to first find the expression for [tex]\((u \circ v)(x)\)[/tex] and then determine its range.
### Step-by-Step Solution:
1. Substitute [tex]\(v(x)\)[/tex] into [tex]\(u(x)\)[/tex]:
The composite function [tex]\((u \circ v)(x)\)[/tex] is defined as [tex]\(u(v(x))\)[/tex]. Substituting [tex]\(v(x) = \frac{1}{x}\)[/tex] into [tex]\(u(x)\)[/tex]:
[tex]\[ u\left(v(x)\right) = u\left(\frac{1}{x}\right) \][/tex]
Using the function [tex]\(u(x) = -2x^2 + 3\)[/tex]:
[tex]\[ u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x}\right)^2 + 3 \][/tex]
2. Simplify the expression:
Simplify the expression inside the composite function:
[tex]\[ u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x^2}\right) + 3 = -\frac{2}{x^2} + 3 \][/tex]
Therefore, the composite function [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (u \circ v)(x) = -\frac{2}{x^2} + 3 \][/tex]
3. Determine the range:
To find the range, analyze the behavior of the function [tex]\(-\frac{2}{x^2} + 3\)[/tex] as [tex]\(x\)[/tex] varies over the real numbers (excluding [tex]\(x = 0\)[/tex], since [tex]\(v(x) = \frac{1}{x}\)[/tex] is undefined at [tex]\(x = 0\)[/tex]).
- As [tex]\(x\)[/tex] approaches 0 from both positive and negative sides, [tex]\(\frac{1}{x^2}\)[/tex] becomes very large, causing [tex]\(-\frac{2}{x^2}\)[/tex] to become a large negative number. Hence, [tex]\(-\frac{2}{x^2} + 3 \)[/tex] approaches [tex]\(-\infty\)[/tex].
- As [tex]\(x\)[/tex] approaches [tex]\(\pm\infty\)[/tex], [tex]\(\frac{1}{x^2}\)[/tex] approaches 0, causing [tex]\(-\frac{2}{x^2}\)[/tex] to approach 0. Therefore, [tex]\(-\frac{2}{x^2} + 3\)[/tex] approaches 3.
In conclusion, the function [tex]\(-\frac{2}{x^2} + 3\)[/tex] can take on any value less than or equal to 3 as [tex]\(x\)[/tex] varies over the real numbers (excluding 0). Therefore, the range of [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (-\infty, 3) \][/tex]
### Final Answer:
The range of [tex]\((u \circ v)(x)\)[/tex] is [tex]\(\boxed{(-\infty, 3)}\)[/tex].
### Step-by-Step Solution:
1. Substitute [tex]\(v(x)\)[/tex] into [tex]\(u(x)\)[/tex]:
The composite function [tex]\((u \circ v)(x)\)[/tex] is defined as [tex]\(u(v(x))\)[/tex]. Substituting [tex]\(v(x) = \frac{1}{x}\)[/tex] into [tex]\(u(x)\)[/tex]:
[tex]\[ u\left(v(x)\right) = u\left(\frac{1}{x}\right) \][/tex]
Using the function [tex]\(u(x) = -2x^2 + 3\)[/tex]:
[tex]\[ u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x}\right)^2 + 3 \][/tex]
2. Simplify the expression:
Simplify the expression inside the composite function:
[tex]\[ u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x^2}\right) + 3 = -\frac{2}{x^2} + 3 \][/tex]
Therefore, the composite function [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (u \circ v)(x) = -\frac{2}{x^2} + 3 \][/tex]
3. Determine the range:
To find the range, analyze the behavior of the function [tex]\(-\frac{2}{x^2} + 3\)[/tex] as [tex]\(x\)[/tex] varies over the real numbers (excluding [tex]\(x = 0\)[/tex], since [tex]\(v(x) = \frac{1}{x}\)[/tex] is undefined at [tex]\(x = 0\)[/tex]).
- As [tex]\(x\)[/tex] approaches 0 from both positive and negative sides, [tex]\(\frac{1}{x^2}\)[/tex] becomes very large, causing [tex]\(-\frac{2}{x^2}\)[/tex] to become a large negative number. Hence, [tex]\(-\frac{2}{x^2} + 3 \)[/tex] approaches [tex]\(-\infty\)[/tex].
- As [tex]\(x\)[/tex] approaches [tex]\(\pm\infty\)[/tex], [tex]\(\frac{1}{x^2}\)[/tex] approaches 0, causing [tex]\(-\frac{2}{x^2}\)[/tex] to approach 0. Therefore, [tex]\(-\frac{2}{x^2} + 3\)[/tex] approaches 3.
In conclusion, the function [tex]\(-\frac{2}{x^2} + 3\)[/tex] can take on any value less than or equal to 3 as [tex]\(x\)[/tex] varies over the real numbers (excluding 0). Therefore, the range of [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[ (-\infty, 3) \][/tex]
### Final Answer:
The range of [tex]\((u \circ v)(x)\)[/tex] is [tex]\(\boxed{(-\infty, 3)}\)[/tex].