SOLUTION: With this problem, I need to find the function and state the domain of each given the following functions f and g. a) (f+g)(x) b) (f-g)(x) c) (f*g)(x) d) (f/g)(x) Probl

Algebra ->  Functions -> SOLUTION: With this problem, I need to find the function and state the domain of each given the following functions f and g. a) (f+g)(x) b) (f-g)(x) c) (f*g)(x) d) (f/g)(x) Probl      Log On


   

Question 22071: With this problem, I need to find the function and state the domain of each given the following functions f and g.
a) (f+g)(x) b) (f-g)(x) c) (f*g)(x) d) (f/g)(x)
Problem:
f(x)=1+1/x; g(x)=1/x
I don't understand how to do this function with the letter "d", I already got "a", "b" and "c".
Can you please help?
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
GOOD THAT YOU ARE ABLE TO DO A,B AND C ..AND WANT HELP ON D...LET US SEE
F(X)=1+1/X=(X+1)/X
G(X)=1/X
WE WANT TO FIND SAY Y =(F/G)(X)=F(X)/G(X)
={(X+1)/X}/(1/X)={(X+1)/X}*X=(X+1)
SO Y=X+1
NOW ON TO FIND DOMAIN : BY THE LOOK OF THIS X+1 CAN HAVE ANY VALUE FOR X IN ALL REAL NUMBERS WE ARE CONCERNED WITH AS Y=X+1 IS DEFINED FOR ALL VALUES OF X..BUT IF WE GO BACK A LITTLE AND SEE HOW WE GOT THIS X+1,WE NOTICE THAT IT WAS OBTAINED BY CANCELLING OUT X ,THAT IS DIVIDING BY X THE NUMERATOR AND DENOMINATOR..SINCE DIVISION BY ZERO IS PROHIBITED,IT MEANS THAT WE WOULD NOT HAVE GOT X+1 WITHOUT DIVISION WITH X ,WHICH IS NOT ALLOWED AT X=0..HENCE THE DOMAIN IS ALL REAL NUMBERS EXCEPT X=0...
THIS CAN BE ALSO GOT FROM ORIGINAL DEFINITION OF Y = F(X)/G(X)...SINCE BOTH THE ORIGINAL FUNCTIONS F(X) AND G(X) INVOLVE DIVISION WITH X AND HENCE ARE NOT DEFINED FOR X=0,OBVIOUSLY IT MEANS THAT Y= F(X)/G(X) IS ALSO NOT DEFINED FOR X=0.