Question

If u(x) = −2x² +3 and v(x)=1/x, what is the range of (uv)(x)?

Answer 1

Given:

$\begin{gathered} u(x)=-2x^2+3 \\ v(x)=\frac{1}{x} \end{gathered}$

Required:

To find the range of the function (uv)(x).

Explanation:

We know that

$\begin{gathered} (uv)(x)=u(v(x)) \\ \\ =u(\frac{1}{x}) \\ \\ =-2(\frac{1}{x^2})+3 \\ \\ =-\frac{2}{x^2}+3 \end{gathered}$

The horizontal asymptote of this function is at y=3.

So, the range of this function is from

$(-\infty,3)$

Final Answer:

The range of (uv)(x) is

$(-\infty,3)$