Question

Question #e4aeb

Answer 1

`D_(f/g)=(-oo,-5)uu(-5,5)uu(5,+oo)`

Explanation:

`D_(f/g) = {AA` `x` `in`R`:` `D_fnnD_g` `-{x:g(x)=0`}`}` `!=Ø`

`D_f=R`
`D_g=R`

`g(x)=0<=>x^2-25=0<=>(x-5)(x+5)=0<=>x=5` or `x=-5`

So `D_(f/g)=R-{-5,5}` `=(-oo,-5)uu(-5,5)uu(5,+oo)`

so for `x` `inR-{-5,5}` :

`(f/g)(x)=f(x)/g(x)=(x+5)/(x^2-25)` `=cancel(x+5)/((x-5)cancel((x+5)))=1/(x-5)`