If u(x)=-2x^2+3 and v(x)=1/x what is the range of (u*v)(x) **the picture should help

we have that
[tex]u(x)=-2 x^{2} +3 \\ \\ v(x)= \frac{1}{x}[/tex]
[tex]u(v(x))=u( \frac{1}{x})=-2( \frac{1}{x} )^{2} +3 \\ u( \frac{1}{x})= \frac{(3 x^{2} -2)}{ x^{2} } [/tex]

using a graph tool
see the attached figure

The horizontal asymptote of this function is at y=3.
So,
the range of this function is from (−∞,3)

Answer Link

MrRoyal MrRoyal · Answer 2 · 2021-09-18T00:03:58+02:00

The range of a function is simply the output values the function can take.

The range of the function [tex]u(v(x))[/tex] is: [tex](-\infty,3)[/tex]

Given

[tex]u(x) = -2x^2 + 3[/tex]

[tex]v(x) = \frac 1x[/tex]

[tex](u\ o\ v)(x)[/tex] can be represented as:

[tex](u\ o\ v)(x) = u(v(x))[/tex]

So, we have:

[tex]u(x) = -2x^2 + 3[/tex]

Substitute v(x) for x

[tex]u(v(x)) = -2(v(x))^2 + 3[/tex]

Substitute [tex]\frac 1x[/tex] for v(x)

[tex]u(v(x)) = -2(\frac 1x)^2 + 3[/tex]

Open brackets

[tex]u(v(x)) = -\frac{2}{x^2} + 3[/tex]

Take LCM

[tex]u(v(x)) = \frac{-2 + 3x^2}{x^2}[/tex]

I've added an attachment of function u(v(x)) to determine its range.

From the attached graph, we have the following observations

The graph has a maximum just below y = 3
The graph opens downwards

This implies that, the range of the function is: [tex](-\infty,3)[/tex] or [tex]-\infty < x < 3[/tex]