At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
could you maybe write this problem out? it looks kind of confusing here :D
Just a sec.
find value of x from denominator that would give you the value of 0.
|dw:1332681829434:dw| That should be the answer, not sure if correct
yes good job!
The problem is that although I have the answer I don't know how to solve it.
the expression under arcsin have to be equal to values that are between -1 and 1
That one I got. But what about the rest?
The value in the square root must be positive.
x^2+5x-6 must be greater than 0
so just put it all togehter
And the denominator may not be zero.
And the final answer would be?
You take the intersection of all the results.
X=[-15;-6)u(1;5] is what I get.
Ok, so the domain for the arcsin is (-15;5), right? Just doing the rest of it...
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.
And finally x+2 is positive whenever x in \[[-2;\infty)\], right?
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?
By the way: replace positive by "non-negative" (including zero) wherever I spoke of square roots, ok?
Will do. Everything should be okay.
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.
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.