If `f(x)=log_(e)(1-x)andg(x)=[x]` then find:
(i) (f+g)(x) (ii) (fg)(x) (iii) `((f)/(g))(x)` (iv) `((g)/(f))(x)`.
Also find `(f+g)(-1),(fg)(0),((f)/(g))(-1),((g)/(f))((1)/(2))`.

Explanation

Clearly, `log_(e)(1-x)` is defined only when `1-xgt0,i.e.,xlt1`.
`:."dom "(f)=(-oo,1)`.
Also, dom (g)=R.
`:."dom "(f)nn"dom "(g)=(-oo,1)nnR=(-oo,1)`
(i) `(f+g)"(-oo,1)toR` is given by
`(f+g)(x)=f(x)+g(x)=log_(e)(1-x)+[x]`.
(ii) `(fg):(-oo,1)toR` is given by
`(fg)(x)=f(x)xxg(x)={log_(e)(1-x)}xx[x]`.
(iii) `{x:g(x)=0}={x:[x]=0}={0,1)`.
`:."dom "((f)/(g))="dom "(f)nn"dom "(g)-{x:g(x)=0}`
`=(-oo,1)nnR-[0,1)=(-oo,0)`.
`:.(f)/(g):(-oo,0)toR` is given by
`((f)/(g))(x)=(f(x))/(g(x))=(log_(e)(1-x))/([x])`.
(iv) `{x:f(x)=0}={x:log_(e)(1-x)=0}={0}`.
`:."dom "((g)/(f))="dom "(g)nn"dom "(f)-{x:f(x)=0}`.
`=Rnn(-oo,1)-{0}=(-oo,0}uu(0,1)`.
`:.(g)/(f):(-oo,0)uu(0,1)toR`
`((g)/(f))(x)=(g(x))/(f(x))=([x])/(log_(e)(1-x))`.
Now, we have:
`(f+g)(-1)=f(-1)+g(-1)=[-1]+log_(e)(1+1)=(log_(e)2)-1`.
`(fg)(0)=f(0)xxg(0)=log_(e)(1-0)xx[0]=(log_(e)1xx0)=(0xx0)=0`.
`((f)/(g))(-1)=(f(-1))/(g(-1))=([-1])/(log_(e)(1+1))=(-1)/(log_(e)2)`.
`((g)/(f))((1)/(2))=g((1)/(2))/(f((1)/(2)))=([(1)/(2)])/(log_(e)(1-(1)/(2)))=([05.])/(log_(e)((1)/(2)))=0`.