Let `f:RtoR:f(x)andg:R-{0}toR:g(x)=(1)/(x)` be the identity function and the reciprocal function respectively.
Now, dom `((f)/(g))="dom "(f)nn"dom "(g)-{x:g(x)=0}`
and `{x:g(x)=0}={x:(1)/(x)=0}=phi`.
`:."dom "((f)/(g))=[RnnR-{0}]-phi=R-{0}`
So, `(f)/(g):R-{0}toR:((f)/(g))(x)=(f(x))/(g(x))=(x)/((1)/(x))=x^(2)`.
Hence, `((f)/(g))(x)=x^(2)"for all "x""inR-{0}`.