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If  `f(x)=x/(x+1)`       and       `g(x)=2/(x^2-1)` 1. (f+g)(x)=? 2. (f-g)(x)=?...

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user4163740 | (Level 1) eNoter

Posted June 19, 2013 at 10:35 AM via web

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If  `f(x)=x/(x+1)`       and       `g(x)=2/(x^2-1)`

1. (f+g)(x)=?

2. (f-g)(x)=?

3. (f*g)(x)=?

4. (f/g)(x)=?


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mjripalda | High School Teacher | (Level 1) Senior Educator

Posted June 19, 2013 at 12:08 PM (Answer #1)

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`f(x)=x/(x+1) `    and  `g(x)= 2/(x^2-1)`

(1) `(f+g)(x)`

To solve, do the operation inside the first parenthesis which is to add the two functions.


`= x/(x+1)+2/(x^2-1)`

To simplify, factor the denominator of g(x).


Now that the denominators are in factored form, determine the LCD.

Note that LCD is the product of all the different factors in the denominators. So, the LCD is (x+1)(x-1). 

`= x/(x+1)*(x-1)/(x-1)+2/((x+1)(x-1))`


`= (x^2-x+2)/((x+1)(x-1))`

Since the numerator is not factorable, then it could no longer be simplified further.

Hence, `(f+g)(x)=(x^2-x+2)/((x+1)(x-1))` .


(2) `(f-g)(x)`

Here, subtract g(x) from f(x).

`= f(x) - g(x)`

`=x/(x+1) - 2/(x^2-1)`

Express these two functions with same denominator using their LCD which is (x+1)(x-1).

`=x/(x+1) - 2/((x+1)(x-1))`




Then, factor the numerator to determine if there is any common factor between the top and bottom


Notice that the factor x+1 appears both at the top and bottom. So cancel it to simplify the expression further.


Hence,  `(f-g)(x)=(x-2)/(x-1)` .


(3) `(f*g)(x)`

Also, apply the operation inside the parenthesis.

`= f(x)*g(x)`


Factor the denominator of g(x) to determine if there are common factors to be cancelled.

`=x/(x+1) * 2/((x+1)(x-1))`

Since there are no factors that appear both at the top and bottom, proceed to multiply straight across.


Hence, `(f*g)(x)=(2x)/((x+1)^2(x-1))` .


(4) `(f/g)(x)`

AS indicated, divide f(x) by g(x).

`= f(x) -: g(x)`

`=x/(x+1) -: 2/(x^2-1)`

Then, apply the steps in dividing fractions.

Flip the second fraction and change the operation from divide to multiply.


To determine if there are factors that can be cancelled, factor x^2-1.


Since (x+1) is a common factor, cancel it.


And then, multiply straight across.


Hence, `(f/g)(x)=(x(x-1))/2` .


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aruv | High School Teacher | (Level 2) Valedictorian

Posted June 19, 2013 at 12:14 PM (Answer #2)

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You question is to find

`(f+g)(x),(f-g)(x) ,(f*g)(x) and (f/g)(x) ` for functions




`Ans 1.`






















`(f/g)(x)=(f(x))/(g(x)) ,g(x)!=0`     for any x





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