SOLUTION: 1. If gof=x^2+2x+1 f(x)= x^2 Find g(x) 2. if g(x)= (x+1)/x gof=2/(x-1) find f(x) Kindly help me on this matter, thank you.

Algebra ->  Rational-functions -> SOLUTION: 1. If gof=x^2+2x+1 f(x)= x^2 Find g(x) 2. if g(x)= (x+1)/x gof=2/(x-1) find f(x) Kindly help me on this matter, thank you.      Log On


   

Question 623114: 1. If gof=x^2+2x+1
f(x)= x^2
Find g(x)
2. if g(x)= (x+1)/x
gof=2/(x-1)
find f(x)
Kindly help me on this matter, thank you.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
1. gof is what g(x) looks like after its x's have been replaced with f(x) (in this case x%5E2). So seeing where the x%5E2 came from is fairly easy. But how dow we replace an x with an x%5E2 and get the x in 2x? With a bit of thought and imagination can figure out that the x%5E2 would turn into an x if it was in a square root. So
g%28x%29+=+x+%2B+2sqrt%28x%29+%2B+1

2. For this one we're given gof and g. And we have to figure out what would you have to replace the x's of g with to get %28x%2B1%29%2Fx? There may well be better ways to figure this out but here is what I did: I replaced the x's of g with f's, set it equal to the given gof, and then I solved for f:
%28f%2B1%29%2Ff+=+2%2F%28x-1%29
Solving for f:
f%2Ff%2B1%2Ff+=+2%2F%28x-1%29
1%2B1%2Ff+=+2%2F%28x-1%29
1%2Ff+=+%282%2F%28x-1%29%29-1
1%2Ff+=+%282%2F%28x-1%29%29-%28x-1%29%2F%28x-1%29
1%2Ff+=+%283-x%29%2F%28x-1%29
f+=+%28x-1%29%2F%283-x%29
Checking:
g%28x%29+=+%28x%2B1%29%2Fx
g%28f%28x%29%29+=+%28%28%28x-1%29%2F%283-x%29%29+%2B+1%29%2F%28%28x-1%29%2F%283-x%29%29

g%28f%28x%29%29+=+%28%28x-1%29+%2B+%283-x%29%29%2F%28x-1%29
g%28f%28x%29%29+=+%282%29%2F%28x-1%29
Check!