Answer:
As the function contains horizontal asymptote y = 3.
Therefore,
[tex]\mathrm{Range\:of\:}\frac{3x^2-2}{x^2}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)<3\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:3\right)\end{bmatrix}[/tex]
Step-by-step explanation:
Considering
[tex]u\left(x\right)=-2x^3+3[/tex]
[tex]v\left(x\right)=\frac{1}{x}[/tex]
We have to determine the range of [tex]\left(u\cdot v\right)\left(x\right)[/tex]
as
[tex]\left(u\circ \:v\right)\left(x\right)=u\left(v\left(x\right)\right)[/tex]
[tex]\:\left(u\circ \:v\right)\left(x\right)=u\left(v\left(x\right)\right)=u\left(\frac{1}{x}\right)[/tex]
[tex]u\left(\frac{1}{x}\right)=-\frac{2}{x^3}+3[/tex]
[tex]u\left(\frac{1}{x}\right)=\frac{3x^2-2}{x^2}[/tex]
As the function contains horizontal asymptote y = 3.
Therefore,
[tex]\mathrm{Range\:of\:}\frac{3x^2-2}{x^2}:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)<3\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:3\right)\end{bmatrix}[/tex]
