Question

How do you solve `((x-1)(3-x))/(x-2)^2>=0`?

Answer 1

`f(x)≥0` for `1≤x≤3` with the exclusion of `x=2` where it is undefined.

Explanation:

The denominator is always positive, so the sign of `f(x)` is determined by the numerator.

The numerator is positive when both factors are positive or both are negative.

`(x-1) >= 0` for `x>=1`
`(3-x) >=0` for `x<=3`

So `f(x) >=0` for `1<=x<=3` with the exclusion of `x=2` where it is undefined.

Answer 2

The answer is `x in [ 1,2[ uu ]2, 3 ]`

Explanation:

Let `f(x)=((x-1)(3-x))/(x-2)^2`

The domain of `f(x)` is `D_f(x)=RR-{2}`

Also the denominator is `>0`, `AAx in D_f(x)`

So, we make the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `1` `color(white)(aaaa)` `3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `-` `color(white)(aaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `3-x` `color(white)(aaaa)` `+` `color(white)(aaa)` `+` `color(white)(aaaa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `-` `color(white)(aaa)` `+` `color(white)(aaaa)` `-`

So,

`f(x)>=0` when `x in [ 1,2[ uu ]2, 3 ]`

graph{((x-1)(3-x))/(x-2)^2 [-12.66, 12.65, -6.33, 6.33]}