posted by MUFFY on .
Determine the domain of the following function: f(x)=sqrt x-7/x^2-5x-6
I think the domain is
[7,all real numbers)
Can you please tell me if this is correct?
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.