You have to be more generous with your parentheses. As is the function does not seem to be properly defined.
It could be
f(x) = sqrt( (x-7)/(x²-5x-6) )
g(x) = sqrt(x-7) / (x²-5x-6) )
I assume it is f(x). If it is g(x), you can proceed along the same lines, and post your results for confirmation.
f(x) can be rewritten as:
f(x) = sqrt( (x-7)/((x-2)(x-3)) )
which tells us that
1. at x<7, the numerator becomes negative.
2. The denominator is a parabola which has zeroes at x=2 and x=3.
3. The denominator is negative between x=2 and x=3. It is positive elsewhere.
From 1 and 3, we conclude that the fraction inside the square radical is
1. negative when x<2,
2. positive when 2<x<3
3. negative when 3<x<7
4. positive when x>7
Also, there are two vertical asymptotes at x=2 and x=3 which should be removed from the domain of f(x).
Therefore the domain of f(x) is:
Actually, there are no parenthesis. The problem is how I wrote it but
sqrt x-7 is over x^2 -5x - 6
I factored it to (x-6) (x+1)
" Actually, there are no parenthesis. The problem is how I wrote it but
sqrt x-7 is over x^2 -5x - 6 "
You have to be more generous with parentheses. Whenever a fraction is transcribed to a single line, you will need to insert parentheses around the numerator AND the denominator to avoid ambiguity.
From what you described, you would write the expression as
sqrt(x-7)/(x^2 -5x - 6)
otherwise it can be (unlikely) interpreted as
sqrt(x) -7/x² -5x -6
which is not a normal function.
Back to the function:
f(x) = sqrt(x-7)/(x^2 -5x - 6)
you have already correctly identified the donmain [7,∞).
The only other considerations required would be to exclude the two singular points (x=2 and x=3) from the domain.
Post your results for checking if you wish.
Thanks I didn't realize how to type it when one is over the other. I will keep it in mind.