Question

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

Answer 1

As the function contains horizontal asymptote y = 3.

Therefore,

`\mathrm{Range\:of\:}(3x^2-2)/(x^2):\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)<3\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:3\right)\end{bmatrix}`

Explanation:

Considering

`u\left(x\right)=-2x^3+3`

`v\left(x\right)=(1)/(x)`

We have to determine the range of
`\left(u\cdot v\right)\left(x\right)`

as

`\left(u\circ \:v\right)\left(x\right)=u\left(v\left(x\right)\right)`

`\:\left(u\circ \:v\right)\left(x\right)=u\left(v\left(x\right)\right)=u\left((1)/(x)\right)`

`u\left((1)/(x)\right)=-(2)/(x^3)+3`

`u\left((1)/(x)\right)=(3x^2-2)/(x^2)`

As the function contains horizontal asymptote y = 3.

Therefore,

`\mathrm{Range\:of\:}(3x^2-2)/(x^2):\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)<3\:\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:3\right)\end{bmatrix}`