I need a little help here-not sure if I got it right:

f(x)= 2/x+1, g(x)= x/x+1

domain of f is {x is part of a set where x >0} and domain of g is {x is part of a set where x is greater than or equal to 0} [0, infinity)U(0,infinity).

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- Sep 28th 2008, 03:56 PMVivianpRange and Domain
I need a little help here-not sure if I got it right:

f(x)= 2/x+1, g(x)= x/x+1

domain of f is {x is part of a set where x >0} and domain of g is {x is part of a set where x is greater than or equal to 0} [0, infinity)U(0,infinity). - Sep 28th 2008, 04:01 PMProve It
OK

Both of the functions are rational functions... in other words, of the form

$\displaystyle f(x) = \frac{g(x)}{h(x)}$, where $\displaystyle g(x)$ and $\displaystyle h(x)$ are continuous functions of $\displaystyle x$.

$\displaystyle f(x)$ is also continuous at all points except where the denominator $\displaystyle (h(x))=0$.

So where does the denominator equal 0? In both cases, where $\displaystyle x = -1$.

So the domain of each function is $\displaystyle \mathbf{R}\backslash \{-1\}$. - Sep 28th 2008, 04:03 PMVivianpRange and domain
The first one...

- Sep 28th 2008, 04:33 PMProve It
Sorry for the double post, didn't realise I edited the first one instead of adding a new one.

OK

Both of the functions are rational functions... in other words, of the form

http://www.mathhelpforum.com/math-he...518d3190-1.gif, where http://www.mathhelpforum.com/math-he...482c35c3-1.gif and http://www.mathhelpforum.com/math-he...e8ba0833-1.gif are continuous functions of http://www.mathhelpforum.com/math-he...155c67a6-1.gif.

http://www.mathhelpforum.com/math-he...a378be62-1.gif is also continuous at all points except where the denominator http://www.mathhelpforum.com/math-he...1881b207-1.gif.

So where does the denominator equal 0? In both cases, where http://www.mathhelpforum.com/math-he...16e275fc-1.gif.

So the domain of each function is $\displaystyle \mathbf{R} \backslash \{-1\}$. - Sep 28th 2008, 04:37 PMVivianprange and domain
wow..so i wasn't even close, huh?

- Sep 28th 2008, 04:42 PMProve It