anonymous
  • anonymous
Find the domain of y=arcsin((x+5)/(10))+(1)/(√(x^(2)+5x-6)-√5:(x+2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
could you maybe write this problem out? it looks kind of confusing here :D
anonymous
  • anonymous
Just a sec.
anonymous
  • anonymous
|dw:1332681628680:dw|

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More answers

anonymous
  • anonymous
find value of x from denominator that would give you the value of 0.
anonymous
  • anonymous
|dw:1332681829434:dw| That should be the answer, not sure if correct
anonymous
  • anonymous
yes good job!
anonymous
  • anonymous
The problem is that although I have the answer I don't know how to solve it.
anonymous
  • anonymous
the expression under arcsin have to be equal to values that are between -1 and 1
anonymous
  • anonymous
That one I got. But what about the rest?
anonymous
  • anonymous
The value in the square root must be positive.
anonymous
  • anonymous
x^2+5x-6 must be greater than 0
anonymous
  • anonymous
so just put it all togehter
anonymous
  • anonymous
x+2>=0
anonymous
  • anonymous
And the denominator may not be zero.
anonymous
  • anonymous
And the final answer would be?
anonymous
  • anonymous
You take the intersection of all the results.
anonymous
  • anonymous
X=[-15;-6)u(1;5] is what I get.
anonymous
  • anonymous
Ok, so the domain for the arcsin is (-15;5), right? Just doing the rest of it...
anonymous
  • anonymous
Okay, thanks.
anonymous
  • anonymous
For the quadratic equation I get the roots -6 and 1. As the function is positive for very small and very large values, values in \[(-\infty;-6] \cup [1,\infty)\] should make the square root positive.
anonymous
  • anonymous
And finally x+2 is positive whenever x in \[[-2;\infty)\], right?
anonymous
  • anonymous
So now the intersection of the three intermediate results should be \[[1;5)\]. This is different from what you got, huh? Did I make a mistake somewhere?
anonymous
  • anonymous
By the way: replace positive by "non-negative" (including zero) wherever I spoke of square roots, ok?
anonymous
  • anonymous
Will do. Everything should be okay.
anonymous
  • anonymous
Sorry, I found one more error in my calc: I think all edge cases are included actually, both in the case of arcsin and in the case of the quadratic equation. (Closed intervals rather than open ones...). So the final result would be [1;5]. Man, I think I am probably just confusing you. Sorry.
anonymous
  • anonymous
Don't worry about confusing me, I'm already confused. But I think I'll just ask my teacher tomorrow. Thank you for helping me and I'll just let you continue with whatever you were doing.

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