Question

Question #e46e8

Answer 1

`(-oo,0)uu[5/3,+oo)`

Explanation:

`"rearrange making x the subject"`

`rArry(3-x^2)=5`

`rArr3y-yx^2=5`

`rArrx^2=(3y-5)/yrArrx=sqrt((3y-5)/y)`

`"now "sqrt((3y-5)/y)>=0`

`3y-5=0rArry=5/3" and "y=0`

`"these values split the range into 3 intervals"`

`(-oo,0),(0,5/3],([5/3,+oo)`

`"choose a "color(blue)"test point in each interval"`

`x=-10tocolor(red)"positive"`

`x=1tocolor(blue)"negative"`

`x=10tocolor(red)"positive"`

`(-oo,0)uu[5/3,+oo)`
graph{5/(3-x^2) [-10, 10, -5, 5]}

Answer 2

The range is `y in (-oo,0) uu[5/3, +oo)`

Explanation:

To find the range, proceed as follows

`y=5/(3-x^2)`

Therefore,

`3-x^2=5/y`

`x^2=3-5/y`

`x^2=(3y-5)/y`

`x=sqrt((3y-5)/y)`

What's under the `sqrt` sign `>=0`

So,

`f(y)=(3y-5)/y>=0`

Solve this inequality with a sign chart,

`color(white)(aaaa)` `y` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `0` `color(white)(aaaaaaaa)` `5/3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `y` `color(white)(aaaaaaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `3y-5` `color(white)(aaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaaa)#`-` `color(white)(aa)` `0` `color(white)(aa)` `+`

`color(white)(aaaa)` `f(y)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `||` `color(white)(aaaa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+`

Therefore,

`f(y)>=0`, `y in (-oo,0) uu[5/3, +oo)`

graph{5/(3-x^2) [-7.9, 7.9, -3.95, 3.95]}