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))`.

If `f(x)=log_(e)(1-x)andg(x)=[x]` then find: (i) (f+g)(x) (ii) (fg)(x)

1.	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))`.
Answer» 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`.