Question #e4aeb

1 Answer
Jim S
Dec 8, 2017

#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)#

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