Domain of sqrt( 19+ x/(9-x))

Homework help.

please help

Domain of sqrt( 19+ x/(9-x))

We need to find the domain of
....__________
.../19 + x/(9-x)
\/

first we know 9-x is not equal to zero
because of the denominator so x is not equal to 9

Next

19 + x/(9-x) has to be greater than or equal to zero

this means

x
▬▬ ≥ -19
9-x

so multiplying by 9-x
on both sides

If 9-x is greater than zero which means -x > -9 so x<9


x≥ -19(9-x)

x≥ -19(9-x)

x≥ -171+19x

-18x≥ -171

so
x≤ 171/18

so
x is less than or equal to 9 9.5

The condition is already x<9 so that is part of the domain


Now if


If 9-x is less than zero which means -x < -9 so x>9

The direction of the inequality reverses when we multiply
by 9-x now

x≤ -19(9-x)

x≤ -19(9-x)

x≤ -171+19x

-18x≤ -171

so
x≥171/18

so
x is greater than or equal to 9.5

The condition is already x>9 so x≥9.5 is part of the domain

The full domain is

(-∞,9)U[9.5,∞)

which is interval notation for ALL real numbers except for
[9,9.5)

Although I'm not sure of what sqrt is, I'm guessing it's square root? But anyways the domain would be x cannot = 9, all real numbers