How do you find #f(g(x))# if #f(x) = (x-3) / (5x+1)# and #g(x) = (x-1) / (x^2)#?

1 Answer
Nov 21, 2015

#f(g(x))=-(3x^2-x+1)/(x^2+5x-5)#

Explanation:

#f(g(x))#, which can also be written as #(f@g)(x)#, simply means that you take #g(x)#, which equals #(x-1)/x^2#, and replace every instance of #x# with that in #f(x)#.

#f(g(x))=((x-1)/x^2-3)/(5((x-1)/x^2)+1)#

Let's try to clear out fractions by multiplying everything by #x^2#.

#f(g(x))=((x-1)/x^2-3)/(5((x-1)/x^2)+1)(x^2/x^2)=(x-1-3x^2)/(5x-5+x^2)#

Neither the numerator nor the denominator can be factored. The terms can be rearranged to be in descending order, but there's not really anything else you can do from here!