#1

I have this info.

f(x) = square root of (x+1)

g(x) = 2/(x-1)

The domain is x<-1 or x>1. How do I arrive at that answer? I know automatically that the domain cannot include x=1 because that would make the denominator in g(x)=0. I know the next step is the plug in 2/(x-1) into f(x), but once I do that, I don't know what to do. I get:

Square root of (all of the following): 2/(x-1) + 1

I tried doing an inequality for finding the domain of this function.

I set the above all greater than or equal to 0 because sq rts have to be greater or equal to zero. I solved that and got x is greater than or equal to -1, which looks fine, but when I plug in 0 into f(g(x)), I get the square root of -1, which doesn't work.

Help please?

f(x) = square root of (x+1)

g(x) = 2/(x-1)

The domain is x<-1 or x>1. How do I arrive at that answer? I know automatically that the domain cannot include x=1 because that would make the denominator in g(x)=0. I know the next step is the plug in 2/(x-1) into f(x), but once I do that, I don't know what to do. I get:

Square root of (all of the following): 2/(x-1) + 1

I tried doing an inequality for finding the domain of this function.

I set the above all greater than or equal to 0 because sq rts have to be greater or equal to zero. I solved that and got x is greater than or equal to -1, which looks fine, but when I plug in 0 into f(g(x)), I get the square root of -1, which doesn't work.

Help please?

#2

Are you just trying to look smart.

Here

E=MC2

Here

E=MC2

#3

why would you plug 0 into f(g(x))? you're making this sound more complicated than it is.

#4

if x < -1, sqrt ( x+1) les on the imaginary plane, so that woulddefinite the other end of your domain. Why the dickens would you plug eqn 1 into eqn 2? Also, the domain is unbounded on the right side, you just have a hole in the function at x=1. if x>1, your funtion still works, but the function is discontinuous at x=1 because the limit from the two directi8ons is different (positive and negative infinity). So, I'm not sure what exactly you're supposed to be finding, but I wouldsay -1<x<(inf.) with a discontinuity at x=1

#5

why would you plug 0 into f(g(x))? you're making this sound more complicated than it is.

To show that my algebra was wrong, so I'm wondering what I did wrong in solving that inequality.

I'm also wondering how to get the right answer, which is x<-1 or x>1.

^Aren't you supposed to plug in g(x) into f(x) and find the domain for

*that*as well?

*Last edited by kirbyrocknroll at Sep 9, 2008,*

#6

You said:

(x) = square root of (x+1)

g(x) = 2/(x-1)

The domain is x<-1 or x>1. ... plug in 2/(x-1) into f(x) ... I get:

Square root of (all of the following): 2/(x-1) + 1

</whatever you said>

By changing 1 to (x-1)/(x-1), you could get sqrt{ [2/(x-1) ] + [ (x-1)/(x-1) ] }, right?

And by combining terms, you could get sqrt[ (x+1)/(x-1) ], right?

Well, by treating 1 and -1 as bounds for the domain, plug in numbers--say, -2, -1, 0, 1, and 2:

f (g (-2) ) = sqrt(1/3), which works.

f (g (-1) ) = sqrt(0), which also works, meaning that the domain includes -1.

f (g (0) ) = sgrt(-1), which is imaginary.

f (g (1) ) prompts division by 0, which can't happen.

f (g (2) ) = sqrt(3), which is also okay.

Therefore, you can say that the domain has to include x>1 and x</= -1.

Good luck with math, man...

(x) = square root of (x+1)

g(x) = 2/(x-1)

The domain is x<-1 or x>1. ... plug in 2/(x-1) into f(x) ... I get:

Square root of (all of the following): 2/(x-1) + 1

</whatever you said>

By changing 1 to (x-1)/(x-1), you could get sqrt{ [2/(x-1) ] + [ (x-1)/(x-1) ] }, right?

And by combining terms, you could get sqrt[ (x+1)/(x-1) ], right?

Well, by treating 1 and -1 as bounds for the domain, plug in numbers--say, -2, -1, 0, 1, and 2:

f (g (-2) ) = sqrt(1/3), which works.

f (g (-1) ) = sqrt(0), which also works, meaning that the domain includes -1.

f (g (0) ) = sgrt(-1), which is imaginary.

f (g (1) ) prompts division by 0, which can't happen.

f (g (2) ) = sqrt(3), which is also okay.

Therefore, you can say that the domain has to include x>1 and x</= -1.

Good luck with math, man...