Linda B. answered • 09/22/16
Tutor
4.9
(9)
Retired High School Math Teacher
Let f(x) = 3/(x+2)∗√(x+5)/√(x+3), Find the domain of f in interval notation.
Domain of f(x), exists anywhere that does not make the denominator 0, or the value under the radical a negative number.
Lets check the zero in the denominator case first.
First of all then, x≠-2, because that would mean that 3/(x+2)=3/(-2+2)=(3/0), which does not exist.
Also, we can not let √(x+3) = 0 for the same reason, so x≠-3, because that would mean that √(x+5)/√(x+3))=√(-3+5)/√(-3+3)=√(2)/√(0) and this just doesn't exist either .
The other thing is that we want a real solution, at least I assume so. Thant means that the value under the radical can not be less than zero.
So we need x + 5 ≥ 0, so x ≥ -5 and we also need x + 3 ≥ 0, so x ≥ -3.
Well since -3 is already not part of the domain due the fact that it makes the denominator 0, we know that x > -3, so that is one end of our interval: (-3,?) From -3, things are pretty ok, until we get to -2, we have a hole there, but then it picks right back up and goes happily off into ∞ with no more problems. So our Domain has two number sets,
(-3, -2) and the other chunk from (-2, ∞), since we need both of those sets of numbers without the intersection, (x=-2), we use the Union symbol.
(-3,-2) U (-2, ∞)
Remember to use the round parenthesis when the boundaries are not included in the Domain, and always with ∞. We could never enclose a concept like infinity, it's not even a number.
Domain of f(x), exists anywhere that does not make the denominator 0, or the value under the radical a negative number.
Lets check the zero in the denominator case first.
First of all then, x≠-2, because that would mean that 3/(x+2)=3/(-2+2)=(3/0), which does not exist.
Also, we can not let √(x+3) = 0 for the same reason, so x≠-3, because that would mean that √(x+5)/√(x+3))=√(-3+5)/√(-3+3)=√(2)/√(0) and this just doesn't exist either .
The other thing is that we want a real solution, at least I assume so. Thant means that the value under the radical can not be less than zero.
So we need x + 5 ≥ 0, so x ≥ -5 and we also need x + 3 ≥ 0, so x ≥ -3.
Well since -3 is already not part of the domain due the fact that it makes the denominator 0, we know that x > -3, so that is one end of our interval: (-3,?) From -3, things are pretty ok, until we get to -2, we have a hole there, but then it picks right back up and goes happily off into ∞ with no more problems. So our Domain has two number sets,
(-3, -2) and the other chunk from (-2, ∞), since we need both of those sets of numbers without the intersection, (x=-2), we use the Union symbol.
(-3,-2) U (-2, ∞)
Remember to use the round parenthesis when the boundaries are not included in the Domain, and always with ∞. We could never enclose a concept like infinity, it's not even a number.
Helen J.
09/22/16