Find the domain of the following functions

- Theloshua

Find the domain of the following functions

- jamiebookeater

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- Theloshua

\[f(x)=\frac{ \sqrt{x+2} }{ x-5 }\]

- Theloshua

\[g(x)=\frac{ x-3 }{ \sqrt{x-1}}\]

- jim_thompson5910

how far did you get with any of these?

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## More answers

- Theloshua

not far at all

- 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

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

agreed so far?

- Theloshua

yes

- 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

x greater or equal to -2

- jim_thompson5910

yes

- Theloshua

oh wait, i already solved for this one. i need help with the second

- jim_thompson5910

so x cannot be 5
AND
x has to be greater than or equal to -2

- 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

what makes the denominator of g(x) equal to 0 ?

- Theloshua

3 cant be an option

- jim_thompson5910

why not?

- Theloshua

because the top

- Theloshua

you subtract -3

- Theloshua

i mean add

- Theloshua

right?

- jim_thompson5910

you can have 0 in the numerator though

- jim_thompson5910

what do you get when you solve sqrt(x-1) = 0 for x?

- Theloshua

1

- 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

also, you need to ensure that x-1 is never negative

- Theloshua

okay

- Theloshua

and the other part?

- jim_thompson5910

solve x-1 >= 0 for x to get what?

- Theloshua

1

- jim_thompson5910

x = 1 ? or x <= 1 ? or x >= 1 ?

- Theloshua

x greater than 1

- jim_thompson5910

greater than or equal to 1, yes

- jim_thompson5910

oh wait, nvm the "or equal to" part

- jim_thompson5910

x cannot equal 1 because it leads to a division by zero error

- Theloshua

okay

- Theloshua

so? explaiiiin im confused tell me how to solve

- jim_thompson5910

you just said it: the domain is x > 1

- jim_thompson5910

any number larger than 1 will work as an input
if x = 1 or smaller, then you run into problems

- Theloshua

okay
so the answer is x is not one and x is greater than 1?

- jim_thompson5910

saying x > 1 implies that x is not equal to 1

- jim_thompson5910

you don't need to say them both if you just say x > 1

- Theloshua

okay, so thats the answer tho

- jim_thompson5910

yes

- Theloshua

thnxxxx

- jim_thompson5910

no problem

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