How do you find the domain and range of f(x)= 2/(x-3)(x+1) I already factored it

Question

tkhunny · Answer

"Determine" might be a better word than "Find".

I presume you mean $f(x) = \dfrac{1}{(x-3)(x+1)}$?  That is not quite qhat you have written.  Remember you Order fo Operations.

Generally, for the Domain, simply start with $\mathbb{R}$ - All Real Numbers.  Then look around for some exceptions that are not included.  In this case, what makes your denominator zero?  Those would be bad.

The Range is a little trickier.  It requires some knowledge of the nature of the function.  In this case, you have two vertical asymptotes.  Both these asymptotes have an odd degree.  This must mean that we are ALMOST at $\mathbb{R}$.  For a rational function, an horizontal asymptote may be an exception.  Can you determine the horizontal asymptote on this one?

anonymous · Answer

For Domain..:-

Since it is a rational Function in the form p(x) / q(x).....q(x) shuld not be equal to 0

so (x-3)(x+1) not equal 0...Find Two Values for x..

anonymous · Answer

that two values makes the function undefined..so. Domian = R - {x1 , x2}

anonymous · Answer

So theres no range?

tkhunny · Answer

How did you decide that?  We had a moment ago, All Real Numbers excepting the location of the Horitzontal Asymptote.  What caused you to believe that it may have disappeared?

anonymous · Answer

@Yahoo!  the domain is R - {x,1 x,2}?

anonymous · Answer

o.O

$$(x-3)(x+1) = 0$$

$$x-3 = 0$$
$$x=3$$

$$x+1 = 0$$

$$x=-1$$

anonymous · Answer

Domain = R-[-1,3}

anonymous · Answer

@Yahoo! How about range? is there not one

anonymous · Answer

@Hero could you help me?