To find the range of the composite function [tex]\((u \circ v)(x)\)[/tex], we need to determine the behavior of [tex]\((u \circ v)(x)\)[/tex] for all values of [tex]\(x\)[/tex] within the domain of [tex]\(v(x)\)[/tex]. Here, we have [tex]\(u(x) = -2x^2 + 3\)[/tex] and [tex]\(v(x) = \frac{1}{x}\)[/tex].
Step-by-step solution:
1. Find the composite function [tex]\( (u \circ v)(x) \)[/tex]:
[tex]\[
(u \circ v)(x) = u(v(x)) = u\left(\frac{1}{x}\right)
\][/tex]
2. Substitute [tex]\( v(x) = \frac{1}{x} \)[/tex] into [tex]\( u(x) \)[/tex]:
[tex]\[
u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x}\right)^2 + 3
\][/tex]
3. Simplify the expression:
[tex]\[
u\left(\frac{1}{x}\right) = -2\left(\frac{1}{x^2}\right) + 3 = -\frac{2}{x^2} + 3
\][/tex]
4. Determine the behavior of [tex]\( -\frac{2}{x^2} + 3 \)[/tex]:
- As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex], [tex]\(\frac{1}{x}\)[/tex] approaches [tex]\(0\)[/tex], so [tex]\(\frac{1}{x^2}\)[/tex] also approaches [tex]\(0\)[/tex], making [tex]\(-\frac{2}{x^2} + 3\)[/tex] approach [tex]\(3\)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\(0\)[/tex] from both positive and negative sides, [tex]\(\frac{1}{x}\)[/tex] approaches [tex]\(\pm \infty\)[/tex], making [tex]\(\frac{1}{x^2}\)[/tex] approach [tex]\(\infty\)[/tex], causing [tex]\(-\frac{2}{x^2}\)[/tex] to approach [tex]\(-\infty\)[/tex]. Thus, [tex]\(-\frac{2}{x^2} + 3\)[/tex] approaches [tex]\(-\infty\)[/tex].
5. Combine the observations to form the range:
- The function approaches the upper bound of [tex]\(3\)[/tex] but never reaches it as [tex]\(x\)[/tex] heads towards [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex].
- The function decreases without bound as [tex]\(x\)[/tex] approaches [tex]\(0\)[/tex].
Therefore, combining these observations, the range of [tex]\((u \circ v)(x)\)[/tex] is:
[tex]\[
(-\infty, 3)
\][/tex]
The correct answer is:
[tex]\[(-\infty, 3)\][/tex]
