Question

For f(x)=(1-x)/(1+x) and g(x)=x/(1-x) , find the simplified form for f [g(x)] and state the domain

Answer 1

[ 1 - (x/(1-x))] / [ 1 +(x/(1-x)) ]

[ 1-x - x]/ [ 1-x +x]

1-2x

if x = 1, g(x) is undefined

Answer 2

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

f(g(x))
= f(x/(1-x) )
= [ 1 - x/(1-x)] / [1 + x/(1-x)]
multiply top and bottom by 1-x
= (1-x - x) / (1 - x + x)
= (1 - 2x)/1
= 1 - 2x

check: pick an unlikely x value and using my calculator
g(2.35) = -1.7407...
f(-1.7407) = -3.699999 or -3.7

using my answer of f(g(x)) = 1-2x
I got -3.7

(my answer is correct)

since in the original, x ≠ ±1
that restriction has to carry through, so

domain is any real number, except x = ±1

Answer 3

Damon and Reiny are both correct about getting to "1 - 2x", but after that they're both utter morons.

You're looking for the domain (x), right? That means you start with all real numbers by default, and have to eliminate values of x that aren't in the domain to narrow it down. That means you need an x-value that sets the denominator to 0, so the result will be undefined (cannot divide by zero).

So, we set the denominator "1 - 2x" equal to 0:

1 - 2x = 0
1 = 2x
1/2 = x

The x-value 1/2 causes the function to be undefined, so we've found our domain: all real numbers EXCEPT x = 1/2.

God, the "helpers" here have gone downhill.

Answer 4

Actually Smart set the numerator equal to 0 which isn’t a restriction. 1 is the only restriction from g(x), since an undefined value cannot be plugged into f(x). G(x) is never -1 (asymptote) so there is no restriction there.