Question

Question #9738d

Answer 1

The domain is `x in ] 10 ,12 [`

Explanation:

What is under the `sqrt` is `>=0`

But the `sqrt` is in the denominator

So,

`sqrt((x-10)(12-x))>0`

Let `f(x)=sqrt((x-10)(12-x))`

Construct the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaaaa)` `10` `color(white)(aaaaaaa)` `12` `color(white)(aaaa)` `-oo`

`color(white)(aaaa)` `x-10` `color(white)(aaaa)` `-` `color(white)(aaaa)` `color(red)(∣∣)` `color(white)(aaa)` `+` `color(white)(aaa)` `color(red)(∣∣)` `color(white)(aa)` `+`

`color(white)(aaaa)` `12-x` `color(white)(aaaa)` `+` `color(white)(aaaa)` `color(red)(∣∣)` `color(white)(aaa)` `+` `color(white)(aaa)` `color(red)(∣∣)` `color(white)(aa)` `-`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `color(red)(∣∣)` `color(white)(aaa)` `+` `color(white)(aaa)` `color(red)(∣∣)` `color(white)(aa)` `-`

Therefore,

`f(x)>0`, when `x in ] 10 ,12 [`