#1

Ok so, Im asking UG for help again regarding math. I cant seem to understand one of the answers to the problem f(x) = sqrt(x)/ sqrt(x^2 - 4x - 5) Im suppose to find the domain of the function the answer from the worksheet said -1 < x =< 0 but i dont get how thats correct. Help?

#2

that answer isn't right...it should be x>5

#3

that answer isn't right...it should be x>5

yeah that was also one of the answer {x > 5 or -1 < x =< 0}

#4

it can't be -1<x=<0, because then you would have a sqrt of a negative number in the numerator, which you can't have if you're dealing with real numbers, which i assume you are

#5

nemesis u beat me 2 it

the bottom factors out to (x-5)(x+1)

However, sinice its a sqrt, the number cant be negative, so x>5 isnt rite.

so -1<x<=0 is rite

the bottom factors out to (x-5)(x+1)

However, sinice its a sqrt, the number cant be negative, so x>5 isnt rite.

so -1<x<=0 is rite

#6

nemesis u beat me 2 it

the bottom factors out to (x-5)(x+1)

However, sinice its a sqrt, the number cant be negative, so x>5 isnt rite.

so -1<x<=0 is rite

nope...

that would be x is negative, which you just said you can't do

#7

nemesis u beat me 2 it

the bottom factors out to (x-5)(x+1)

However, sinice its a sqrt, the number cant be negative, so x>5 isnt rite.

so -1<x<=0 is rite

how is x > 5 not right? if I plug in something like 6 it would make the function equal to a postive number on both the numerator and denominator

#8

**** wut was i thinking

#9

so can anyone explain to me how this makes sense...?

#10

whats wrong with x>5? if you substitute in x=1000, you get sqrt(1000)/sqrt(995*1001), which is a valid number (altho not a pretty one)

so x>5 is part of the domain, but x has to be greater than 0 because of the numerator, which is straight up sqrt(x) -> x has to be positive because of that part. so i would think that x>5 is your domain, period. the back of the book has been known to be wrong, if that's where your getting the answer.

well now that weve got that settled, to actually do it, you find all the ways to make the function invalid, such as a negative under a root or a zero in teh denominator. then set up equations to trigger those exact results, and figger out what values of x you can and cannot use.

so x>5 is part of the domain, but x has to be greater than 0 because of the numerator, which is straight up sqrt(x) -> x has to be positive because of that part. so i would think that x>5 is your domain, period. the back of the book has been known to be wrong, if that's where your getting the answer.

well now that weve got that settled, to actually do it, you find all the ways to make the function invalid, such as a negative under a root or a zero in teh denominator. then set up equations to trigger those exact results, and figger out what values of x you can and cannot use.

*Last edited by Mr_BTP at Oct 6, 2007,*

#11

the -1 < x =< 0 is wrong...it has to be a mistake in the answer key.

as an example, if you plugged in 0 to the equation, you would get 0/sqrt(-5), which you can't do because that would be a square root of a negative number. the correct answer is x>5

as an example, if you plugged in 0 to the equation, you would get 0/sqrt(-5), which you can't do because that would be a square root of a negative number. the correct answer is x>5

#12

You need x >= 0 So the numerator isn't imaginary.

You also need x^2 - 4x - 5 >= 0 so the denominator isn't imaginary.

So the domain has to be >= 0 (Due to first constraint). And x^2 -4x - 5 >= 0 for x <= -1 and x >= 5. So with both our constaints, we only have x >= 5.

However, if both the numerator and denominator are imaginary, then f(x) ends up being real. Both the numerator and denominator are imaginary for -1 < x < 0.

So domain is x >= 5 and -1 < x < 0. I'm not sure why there is the equality to 0 in your solutions..maybe i've done something wrong.

You also need x^2 - 4x - 5 >= 0 so the denominator isn't imaginary.

So the domain has to be >= 0 (Due to first constraint). And x^2 -4x - 5 >= 0 for x <= -1 and x >= 5. So with both our constaints, we only have x >= 5.

However, if both the numerator and denominator are imaginary, then f(x) ends up being real. Both the numerator and denominator are imaginary for -1 < x < 0.

So domain is x >= 5 and -1 < x < 0. I'm not sure why there is the equality to 0 in your solutions..maybe i've done something wrong.

#13

You need x >= 0 So the numerator isn't imaginary.

You also need x^2 - 4x - 5 >= 0 so the denominator isn't imaginary.

So the domain has to be >= 0 (Due to first constraint). And x^2 -4x - 5 >= 0 for x <= -1 and x >= 5. So with both our constaints, we only have x >= 5.

However, if both the numerator and denominator are imaginary, then f(x) ends up being real. Both the numerator and denominator are imaginary for -1 < x < 0.

So domain is x >= 5 and -1 < x < 0. I'm not sure why there is the equality to 0 in your solutions..maybe i've done something wrong.

oh i forgot about that whole thing about I cancelling out itself.........

disregard most of my prior post - this guy's right. i didnt think this one through all the way

#14

You need x >= 0 So the numerator isn't imaginary.

You also need x^2 - 4x - 5 >= 0 so the denominator isn't imaginary.

So the domain has to be >= 0 (Due to first constraint). And x^2 -4x - 5 >= 0 for x <= -1 and x >= 5. So with both our constaints, we only have x >= 5.

However, if both the numerator and denominator are imaginary, then f(x) ends up being real. Both the numerator and denominator are imaginary for -1 < x < 0.

So domain is x >= 5 and -1 < x < 0. I'm not sure why there is the equality to 0 in your solutions..maybe i've done something wrong.

How were you able to get x <= -1?

#15

probably shouldn't be getting into a math debate when i'm drunk but...

i have to disagree with nova lau

if you're dealing with only real numbers, you can't have any imaginary numbers in the solution set. so while the imaginary numbers would cancel out, you wouldn't be able to get them in the first place if you are in the set of real numbers. therefore, there shouldn't be any talk about x being negative, since it would cause the numerator to be negative. also, x can't be equal to 5, it can only be greater than, since it would cause the denominator to be 0 if it was equal to 5.

i have to disagree with nova lau

if you're dealing with only real numbers, you can't have any imaginary numbers in the solution set. so while the imaginary numbers would cancel out, you wouldn't be able to get them in the first place if you are in the set of real numbers. therefore, there shouldn't be any talk about x being negative, since it would cause the numerator to be negative. also, x can't be equal to 5, it can only be greater than, since it would cause the denominator to be 0 if it was equal to 5.

#16

When -1 < x < 0, neither f(x) nor x are imaginary. The domain ask for the set of values for x where f(x) is defined and real. f(x) is defined and real for -1 < x < 0, thats all that matters.

However you're right about x > 5, not x >= 5.

However you're right about x > 5, not x >= 5.

#17

When -1 < x < 0, neither f(x) nor x are imaginary. The domain ask for the set of values for x where f(x) is defined and real. f(x) is defined and real for -1 < x < 0, thats all that matters.

However you're right about x > 5, not x >= 5.

how is it not imaginary? If you plug in something like -.5 itll make both the numerator and denominator negative square roots which cant happen.

#18

when -1<x<0, f(x) is imaginary. for example, lets say x is -0.5.

then you would have, in the numerator, the sqrt of -0.5. the answer to that is imaginary, and therefore a negative number can't be part of the domain.

then you would have, in the numerator, the sqrt of -0.5. the answer to that is imaginary, and therefore a negative number can't be part of the domain.

#19

when -1<x<0, f(x) is imaginary. for example, lets say x is -0.5.

then you would have, in the numerator, the sqrt of -0.5. the answer to that is imaginary, and therefore a negative number can't be part of the domain.

exactly so I still dont understand how -1 < x <= 0 is factual answer to this problem

#20

exactly so I still dont understand how -1 < x <= 0 is factual answer to this problem

it's not it's a mistake in the answer key