f (g(x) ) = (1 + g(x) )/(1 - g(x) )
= (1 + x/(1-x) / (1 - x/(1-x) )
multiply top and bottom by (1-x)
= (1 - x + x) / (1 - x - x)
= 1/(1 - 2x)
check:
let x = 2
g(2) = 2/(1-2) = -2
f(g(2)) = f(-2)
= (1-2)/(1+2) = -1/3
in mine f(g(2)) = 1/(1-4) = -1/3
it is "highly likely" that my answer is correct
f(x)=(1+x)/(1-x),g(x)=x/(1-x), find the simplest form of f[g(x)] and state the domain.
My work:
(1+ (x/(1-x)))/(1-(x/(1-x)))
(1+x)/(1-x)/(1-x)/(1-x)
2 answers
Reiny's answer is correct, according to
(1 + x/(1-x) / (1 - x/(1-x) )
The domain is all reals except x = 1/2 or 1
Even though the domain of 1/(1-2x) only excludes x = 1/2, we also need to exclude x=1 since f(1) and g(1) are undefined, so f(g(1)) cannot be defined either.
(1 + x/(1-x) / (1 - x/(1-x) )
The domain is all reals except x = 1/2 or 1
Even though the domain of 1/(1-2x) only excludes x = 1/2, we also need to exclude x=1 since f(1) and g(1) are undefined, so f(g(1)) cannot be defined either.
1 answer