Theloshua
  • Theloshua
Find the domain of the following functions
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Theloshua
  • Theloshua
\[f(x)=\frac{ \sqrt{x+2} }{ x-5 }\]
Theloshua
  • Theloshua
\[g(x)=\frac{ x-3 }{ \sqrt{x-1}}\]
jim_thompson5910
  • jim_thompson5910
how far did you get with any of these?

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More answers

Theloshua
  • Theloshua
not far at all
Theloshua
  • Theloshua
like i need helpppp i haves so much work to do and so lil time, i dont wanna be up all nite...
jim_thompson5910
  • 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
jim_thompson5910
  • jim_thompson5910
agreed so far?
Theloshua
  • Theloshua
yes
jim_thompson5910
  • 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
Theloshua
  • Theloshua
x greater or equal to -2
jim_thompson5910
  • jim_thompson5910
yes
Theloshua
  • Theloshua
oh wait, i already solved for this one. i need help with the second
jim_thompson5910
  • jim_thompson5910
so x cannot be 5 AND x has to be greater than or equal to -2
jim_thompson5910
  • jim_thompson5910
same idea with g(x) you cannot divide by 0 and you cannot take the square root of a negative number
jim_thompson5910
  • jim_thompson5910
what makes the denominator of g(x) equal to 0 ?
Theloshua
  • Theloshua
3 cant be an option
jim_thompson5910
  • jim_thompson5910
why not?
Theloshua
  • Theloshua
because the top
Theloshua
  • Theloshua
you subtract -3
Theloshua
  • Theloshua
i mean add
Theloshua
  • Theloshua
right?
jim_thompson5910
  • jim_thompson5910
you can have 0 in the numerator though
jim_thompson5910
  • jim_thompson5910
what do you get when you solve sqrt(x-1) = 0 for x?
Theloshua
  • Theloshua
1
jim_thompson5910
  • jim_thompson5910
so if x = 1, then the whole denominator turns to 0 which is why x = 1 is NOT part of the domain
jim_thompson5910
  • jim_thompson5910
also, you need to ensure that x-1 is never negative
Theloshua
  • Theloshua
okay
Theloshua
  • Theloshua
and the other part?
jim_thompson5910
  • jim_thompson5910
solve x-1 >= 0 for x to get what?
Theloshua
  • Theloshua
1
jim_thompson5910
  • jim_thompson5910
x = 1 ? or x <= 1 ? or x >= 1 ?
Theloshua
  • Theloshua
x greater than 1
jim_thompson5910
  • jim_thompson5910
greater than or equal to 1, yes
jim_thompson5910
  • jim_thompson5910
oh wait, nvm the "or equal to" part
jim_thompson5910
  • jim_thompson5910
x cannot equal 1 because it leads to a division by zero error
Theloshua
  • Theloshua
okay
Theloshua
  • Theloshua
so? explaiiiin im confused tell me how to solve
jim_thompson5910
  • jim_thompson5910
you just said it: the domain is x > 1
jim_thompson5910
  • jim_thompson5910
any number larger than 1 will work as an input if x = 1 or smaller, then you run into problems
Theloshua
  • Theloshua
okay so the answer is x is not one and x is greater than 1?
jim_thompson5910
  • jim_thompson5910
saying x > 1 implies that x is not equal to 1
jim_thompson5910
  • jim_thompson5910
you don't need to say them both if you just say x > 1
Theloshua
  • Theloshua
okay, so thats the answer tho
jim_thompson5910
  • jim_thompson5910
yes
Theloshua
  • Theloshua
thnxxxx
jim_thompson5910
  • jim_thompson5910
no problem

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