# 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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`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.

`=f(x)+g(x)`

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

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

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

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)/((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))`

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

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

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

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

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

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

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

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

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

Also, apply the operation inside the parenthesis.

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

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

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.

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

**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.

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

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

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

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

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

And then, multiply straight across.

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

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

You question is to find

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

`f(x)=x/(x+1)`

`and`

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

`Ans 1.`

`(f+g)(x)=f(x)+g(x)`

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

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

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

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

Ans2.

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

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

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

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

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

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

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

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

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

`Ans3.`

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

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

`=(2x)/(x^3-x+x^2-1)`

`=(2x)/(x^3+x^2-x-1)`

`Ans4.`

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

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

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

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

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