What is the domain of #f(x) = ((x+34)(x-19))/(9(x+24)(x+34))#?

The (maximum) domain of a function is given by all the values #x# such that #f(x)# is well defined. For example, in the function

#f(x) = 1/x#

#f# is not defined at 0, because #f(0) = 1//0#, meaning we would have to divide by 0. As such, the domain of #1//x# is

#1/x : RR"\"{0}#

Where "\" represents substraction of sets.

We have the function

#f(x) = ((x+34)(x-19))/(9(x+24)(x+34))#

It might be tempting to cancel out the #(x+34)# and solve from there; however, this would be erroneous. If #x# takes the value #-34#, the function is undefined; we cannot claim that the following is true:

#(-34+34)/(-34+34) = 1#

By doing this, you can (falsely) prove that 0/0=2.

As such, we can redefine #f#:

#f(x) = {((x-19)/(9(x+24)) " if " x!=34),("undefined if " x=34) :}#

Again, if #x=-24#, #f# is undefined:

#f(color(blue)(-24)) = (color(blue)(-24)-19)/(9(color(blue)(-24)+24))=(-43)/0#

Hence, #f# is undefined at #x=-24# and #x=-34#. The domain of #f# is

#f:RR"\"{-34,-24}#

The only thing that will make this expression undefined is when the denominator is zero, so let's set it to zero to find the excluded values:

#x+24=0=>x=-24#

#x+34=0=>x=-34#

Therefore, the values #x=-24# and #-34# will make this expression undefined, so we can say the domain is

#x in RR, x!= -24, -34#

Hope this helps!