# How do you solve functions that have fractions?If f(x)=1/x and g(x)=x/1+x, how would you find each of the following: (f+g)(x), (f-g)(x), (f*g)(x), (f/g)(x), (g.f)(x), and (f.g)(x) I don't know...

How do you solve functions that have fractions?

If f(x)=1/x and g(x)=x/1+x, how would you find each of the following:

(f+g)(x), (f-g)(x), (f*g)(x), (f/g)(x), (g.f)(x), and (f.g)(x)

I don't know where to begin when approaching functions with fractions in them. Please help me!

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### 1 Answer

(f+g)(x) is like saying f(x) + g(x); (f/g)(x) is like saying f(x)/g(x), and so on. (g.f)(x) is like saying g(f(x)).

So,

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

To add fractions, make the denominators equal, like this:

1/x + x/(1+x) = (1/x)*(1 + x)/(1 + x) + x/(1 + x)*(x/x)

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

See how the denominators are now the same? So we can combine:

(f+g)(x)= **(1 + x + x^2)/(x + x^2)**

(f-g)(x) is the same, just with a minus sign: f(x) - g(x):

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

(f*g)(x) = f(x)*g(x) = (1/x)*x/(1+x) = **1/(1+x)**

(f/g)(x) = f(x)/g(x) = (1/x)/(x/(1+x)) = (1/x)*(1+x)/x = **(1+x)/x^2**

(g.f)(x) = g(f(x)) = f(x)/(1 + f(x) ) = (1/x)/(1 + 1/x ) = 1/(x(1 + 1/x )) = **1/(x + 1)**

(f.g)(x) = f(g(x)) = 1/g(x) = (x+1)/x = **1 + 1/x**