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Mathematics 17 Online
OpenStudy (anonymous):

What is the domain of sqrt x^2+2x-24/x^2-9

OpenStudy (anonymous):

\[\sqrt{x^+2x-24/x^2-9}\]

OpenStudy (anonymous):

You want to find the numbers/intervals for which the function is valid. For rationals you cannot have zero in the denominator and for square root functions, you can not have a negative number.

OpenStudy (anonymous):

So infinity to infinity?

OpenStudy (anonymous):

Not quite. You need to solve \[x^2-9\neq0\] for x. And then \[\sqrt{(x^2+2x-24)/(x^2-9)}\ge0\] for x.

OpenStudy (anonymous):

x=3 for the first one and x= -6 x= -4/ x= -3 x=3

OpenStudy (anonymous):

So the domain is \[x \neq \pm3 and -6\ge x \ge 4\]

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