find the domain of the function of "f(x)=sqrt of x+9/(x+6)(X+4)"?

Answer 1

The restrictions to keep in mind in this question are (1) restrictions of the square root, and (2) restrictions of the fraction.

Restrictions of the square root: what's inside the square root must be greater than or equal to 0. Restrictions on the fraction: the denominator can't be 0.

So if we take the first case, sqrt(x+9), the inside cannot be negative, so x+9 >= 0, or

x >= -9

Let's use that as our first guidelines. Second guideline is that the denominator, (x+6)(x+4) cannot equal zero. So we calculate when they ARE equal to 0, and get x = -6 and x = -4. This means x is NOT equal to -6 and x is NOT equal to -4

So we have
x >= -9
x != -4
x != -6

We combine all of those to get the domain of the function to be

[-9 to -6) union (-6, -4) union (-4, infinity)

Answer 2

Two big problems with domain in the real number are dividing by zero and taking the square root of a negative number. Neither are allowed in the real number system.
The particular answer to your question depends on whether you mean sqrt((x+9)/(x+6)(x+4)) or (sqrt (x+9))/(x+6)(x+4)

In both cases your domain cannot include either -6 or -4. Either of these answers would lead to a divide by zero kind of error.

In the latter case, it is simply (x+9) which must be positive. That is because (x+9) is in the square root and you are not allowed to take a square root of a negative number. In this case, x must be greater than or equal to -9, since anything less than that would lead to a negative square root. In this case, your answer would be "x is greater than or equal to -9, but not equal to -6 or -4"

In the former case, it is a little more tricky to see what is a negative and what is a positive square root, since (x+9)/(x+6)(x+4) takes on variously positive and negative values a lot. When x is less than -9, all three values are negative so the total is negative (not allowed). Between -9 and -6, only two of the values are negative, yielding a positive (allowed). Between -6 and -4, only one value is negative, yielding a negative (not allowed). When x is greater than -4, all three are positive, yielding a positive (allowed). Therefore your answer would be. "x is greater and or equal to -9 and less than -6 OR x is greater than -4"

Answer 3

The domain is all real numbers not equal to -6 and -4. You want to set your denominator to zero and solve. So you would be solving x+6=0 and get that x=-6, do the same for x+4=0 and get x=-4. Being that you cannot have a zero in the denominator you would know that your domain is all real numbers not equal to -6 and all real numbers not equal to -4.

Oh yea, I forgot about the square root...the person above me is correct.

Answer 4

We can find domaine of f(x)=sqrt x+9/(x+6)(x+4) if (x+9)/(x+6)(x+4) > = 0 (superior or equa to zero)
x+9>=0 and set (x+4)(x+6) =0
x > =-9 ; x belongs to intervall -9 to the plus infini for (x+9)
x'=-6 and x''=-4 belong to the intervall less infini to plus infini and f(x) is not.
Then,domaine of the function of f(x) belongs to the intervall -9,-6,-4 to the plus infini.
Df(x)=[-9,-6[u]-6,-4[u]-4,plus infini[