Question

Find the domain of the function defined by

Answer 1

\[f(x)=\sqrt{x^2-5x-6}\]
Please show all work.

Answer 2

It depends on if you're limiting yourself to the real number system or not. Any idea?

Answer 3

Well the question doesn't state any limits. ): I honestly don't even know how to start.

Answer 4

Well, looking at it kinda superficially (it's been awhile since I've done this stuff, maybe someone will come along who knows better) we have to do some basic algebra for starters.

So, let's factor the polynomial.

\[f(x) = \sqrt{(x-6)(x+1)}\]

Now, I'm sure there's gotta be some classy way that I'm forgetting to make this inequality actually mathematically proved, but we know the function is undefined if we root a negative number (at least if we're limiting to the real number system), or if we divide by zero. There's no division here, so I'm assuming they're saying that negative roots are undefined.

So, when is the stuff under the sqrt symbol negative?

Answer 5

you're doing it right...your method is correct

Answer 6

Uhh... I don't know. :/ I will probably fail my exam tomorrow.

Answer 7

It's an inequality of some kind...

Answer 8

what inequality makes a square root expression undefined?
It is undefined when the expression inside the square root is negative

Answer 9

\[(-\infty,-1)\cup(6,\infty)\]

See if you can work backwards from here.

Answer 10

I knew the answer beforehand. I just didn't know how to do it. :P

Answer 11

Alright, terrible drawing but maybe it will help!

When we set x < 6 and x < -1 (at which point we know the individual pieces of the polynomials shift from 0 to negative) we can see when the two are either negative, positive, or the same. When they're the same, they're positive. When they're different, they're negative. By lining them up top to bottom we can see when they're alike and different, and thus know when the number beneath the root symbol will be negative. In so doing, we can determine when the function is defined, and by looking at what isn't defined what is therefore undefined.

Any clearer?

Answer 12

So, negative inifinite up to but NOT INCLUDING -1, and then it becomes negative, and then from 6 but NOT INCLUDING 6 to positive infinite it remains positive and defined: thus (-oo, -1) U (6, oo)