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OpenStudy (theloshua):
\[f(x)=\frac{ \sqrt{x+2} }{ x-5 }\]
OpenStudy (theloshua):
\[g(x)=\frac{ x-3 }{ \sqrt{x-1}}\]
jimthompson5910 (jim_thompson5910):
how far did you get with any of these?
OpenStudy (theloshua):
not far at all
OpenStudy (theloshua):
like i need helpppp i haves so much work to do and so lil time, i dont wanna be up all nite...
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jimthompson5910 (jim_thompson5910):
well first off, hopefully you see that x-5 in the denominator cannot be zero, so x-5 = 0 leads to x = 5
if x = 5, then the denominator x-5 is 0
that means x = 5 is NOT part of the domain
jimthompson5910 (jim_thompson5910):
agreed so far?
OpenStudy (theloshua):
yes
jimthompson5910 (jim_thompson5910):
in the numerator of f(x), we have a radical
you cannot take the square root of a negative number, so x+2 cannot be negative
x+2 is either 0 or positive
solve \[\Large x+2 \ge 0\] for x. Tell me what you get
OpenStudy (theloshua):
x greater or equal to -2
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jimthompson5910 (jim_thompson5910):
yes
OpenStudy (theloshua):
oh wait, i already solved for this one. i need help with the second
jimthompson5910 (jim_thompson5910):
so x cannot be 5
AND
x has to be greater than or equal to -2
jimthompson5910 (jim_thompson5910):
same idea with g(x)
you cannot divide by 0 and you cannot take the square root of a negative number
jimthompson5910 (jim_thompson5910):
what makes the denominator of g(x) equal to 0 ?
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OpenStudy (theloshua):
3 cant be an option
jimthompson5910 (jim_thompson5910):
why not?
OpenStudy (theloshua):
because the top
OpenStudy (theloshua):
you subtract -3
OpenStudy (theloshua):
i mean add
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OpenStudy (theloshua):
right?
jimthompson5910 (jim_thompson5910):
you can have 0 in the numerator though
jimthompson5910 (jim_thompson5910):
what do you get when you solve sqrt(x-1) = 0 for x?
OpenStudy (theloshua):
1
jimthompson5910 (jim_thompson5910):
so if x = 1, then the whole denominator turns to 0
which is why x = 1 is NOT part of the domain
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jimthompson5910 (jim_thompson5910):
also, you need to ensure that x-1 is never negative
OpenStudy (theloshua):
okay
OpenStudy (theloshua):
and the other part?
jimthompson5910 (jim_thompson5910):
solve x-1 >= 0 for x to get what?
OpenStudy (theloshua):
1
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jimthompson5910 (jim_thompson5910):
x = 1 ? or x <= 1 ? or x >= 1 ?
OpenStudy (theloshua):
x greater than 1
jimthompson5910 (jim_thompson5910):
greater than or equal to 1, yes
jimthompson5910 (jim_thompson5910):
oh wait, nvm the "or equal to" part
jimthompson5910 (jim_thompson5910):
x cannot equal 1 because it leads to a division by zero error
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OpenStudy (theloshua):
okay
OpenStudy (theloshua):
so? explaiiiin im confused tell me how to solve
jimthompson5910 (jim_thompson5910):
you just said it: the domain is x > 1
jimthompson5910 (jim_thompson5910):
any number larger than 1 will work as an input
if x = 1 or smaller, then you run into problems
OpenStudy (theloshua):
okay
so the answer is x is not one and x is greater than 1?
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jimthompson5910 (jim_thompson5910):
saying x > 1 implies that x is not equal to 1
jimthompson5910 (jim_thompson5910):
you don't need to say them both if you just say x > 1
OpenStudy (theloshua):
okay, so thats the answer tho
jimthompson5910 (jim_thompson5910):
yes
OpenStudy (theloshua):
thnxxxx
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