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 understand what happens when we substitute [tex]\(v(x)\)[/tex] into [tex]\(u(x)\)[/tex].
Let's find [tex]\(u \left( v(x) \right)\)[/tex]:
1. Start with the definition: [tex]\( u(v(x)) = u \left( \frac{1}{x} \right) \)[/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]
Now we have the composite function:
[tex]\[
(u \circ v)(x) = -\frac{2}{x^2} + 3
\][/tex]
Next, we determine the range of [tex]\((u \circ v)(x)\)[/tex].
4. Notice that [tex]\(\frac{1}{x^2}\)[/tex] is always positive for any [tex]\(x \neq 0\)[/tex]. Therefore, [tex]\(\frac{2}{x^2}\)[/tex] is always positive, making [tex]\(-\frac{2}{x^2}\)[/tex] always negative.
5. As [tex]\(\frac{1}{x^2}\)[/tex] becomes large (i.e., as [tex]\(x\)[/tex] moves away from 0 in either positive or negative direction), [tex]\(-\frac{2}{x^2} \)[/tex] approaches 0 from the negative side. Therefore, as [tex]\(x\)[/tex] approaches [tex]\(\infty\)[/tex] or [tex]\(-\infty\)[/tex], [tex]\(-\frac{2}{x^2}\)[/tex] approaches 0. Adding 3 to this value, we get an expression that approaches 3.
6. However, since [tex]\(-\frac{2}{x^2}\)[/tex] is always negative, [tex]\(-\frac{2}{x^2} + 3\)[/tex] will always be less than 3.
We conclude that the function [tex]\((u \circ v)(x) = -\frac{2}{x^2} + 3\)[/tex] can never quite reach 3, but can get arbitrarily close to it from below. At the other end, as [tex]\(x\)[/tex] approaches 0, [tex]\(-\frac{2}{x^2}\)[/tex] dominates and the expression can become arbitrarily negative.
Therefore, the range of [tex]\((u \circ v)(x)\)[/tex] is [tex]\((-∞, 3)\)[/tex].
Thus, the correct answer is:
[tex]\[
(-\infty, 3)
\][/tex]
