Question

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

Answer 1

`"Domain"= RR"\"{-34,-24}`

Explanation:

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}`

Answer 2

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

Explanation:

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!