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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    \[\sqrt{x^+2x-24/x^2-9}\]

  2. anonymous
    • 5 years ago
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    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.

  3. anonymous
    • 5 years ago
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    So infinity to infinity?

  4. anonymous
    • 5 years ago
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    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.

  5. anonymous
    • 5 years ago
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    x=3 for the first one and x= -6 x= -4/ x= -3 x=3

  6. anonymous
    • 5 years ago
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    So the domain is \[x \neq \pm3 and -6\ge x \ge 4\]

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