How do you find the domain of #g(x) = x/((x-1)(x+3))#?

1 Answer
Shura
May 12, 2015

The answer is : #D = RR-{-3;1}#.

The domain of a function is all the values that #x# can take.

Think about it this way : In a usual function, it is all the real numbers (#RR#) except some numbers or ranges of numbers. Basically, it is all the #x# you can't use because your function wouldn't give you a finite result (it would give you #oo# or something you can't calculate, like #sqrt(-3)#).

So we have : #g(x) = x/((x-1)(x+3))#

In your case, since we have a fraction, you know you can't divide by zero.

So you will avoid having #-3# and #1# as values of #x# and exclude them from the domain :

#D = RR-{-3;1}# or

#D = ]-oo;-3[ uu ]-3;1[ uu ]1; +oo[#.

You got your answer.

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