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September 17, 2014

September 17, 2014

Posted by **MUFFY** on Monday, September 21, 2009 at 8:41pm.

I think the domain is

[7,all real numbers)

Can you please tell me if this is correct?

- PRE-CALC -
**MathMate**, Monday, September 21, 2009 at 11:49pmYou 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) )

or

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:

(2,3)∪(7,∞)

- PRE-CALC -
**MUFFY NEEDS MORE HELP**, Tuesday, September 22, 2009 at 5:32pmActually, 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)

- PRE-CALC -
**MathMate**, Tuesday, September 22, 2009 at 5:40pm" 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)

=sqrt(x-7)/((x-2)(x-3))

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.

- PRE-CALC -
**MUFFY NEEDS MORE HELP**, Tuesday, September 22, 2009 at 11:44pmThanks I didn't realize how to type it when one is over the other. I will keep it in mind.

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