Given:
[tex]\begin{gathered} u(x)=-2x^2+3 \\ v(x)=\frac{1}{x} \end{gathered}[/tex]Required:
To find the range of the function (uv)(x).
Explanation:
We know that
[tex]\begin{gathered} (uv)(x)=u(v(x)) \\ \\ =u(\frac{1}{x}) \\ \\ =-2(\frac{1}{x^2})+3 \\ \\ =-\frac{2}{x^2}+3 \end{gathered}[/tex]The horizontal asymptote of this function is at y=3.
So, the range of this function is from
[tex](-\infty,3)[/tex]Final Answer:
The range of (uv)(x) is
[tex](-\infty,3)[/tex]