possible solutions
1/x
(1 + x)/x
(1 - x)/(1 - 2x)
1 - x
or is it none of these
2007-01-27 02:08:46 · 5 answers · asked by Olivia 4 in Science & Mathematics ➔ Mathematics
f(x) = 1 / (1 - x)
g(x) = x / (1 - x)
h(x) = 1 / (1 + x)
Therefore,
f(g(x)) = f( x / (1 - x) )
And we plug in x/(1 - x) for every occurrance of x in f(x).
f(x/(1 - x)) = 1 / (1 - [x/(1 - x)])
This is a complex fraction, so we should make it into a simple fraction by multiplying top and bottom by (1 - x). This gives us
(1 - x) / [ (1 - x) - x]
This simplifies to
(1 - x) / (1 - 2x)
2007-01-27 02:14:44 · answer #1 · answered by Puggy 7 · 1⤊ 0⤋
f(x) = 1/(1 - x), g(x) = x/(1 - x),
f[g(x)] =1/[1-g(x)] = 1/[1- x/(1-x)] = (1-x)/ (1-2x) Answer
2007-01-27 10:15:30 · answer #2 · answered by Anonymous · 1⤊ 0⤋
we have
f(x) = 1/(1 - x), g(x) = x/(1 - x),
so f[g(x)] =1/[1-g(x)]
= 1/[1- x/(1-x)]
= 1/ [ (1-2x) / (1-x) ]
= (1-x)/(1-2x) .ANSWER
2007-01-27 10:20:02 · answer #3 · answered by rajeev_iit2 3 · 1⤊ 0⤋
the ans. is (1-x)/(1-2x)
2007-01-27 10:16:42 · answer #4 · answered by angel_4_life_88_2002 1 · 1⤊ 0⤋
f(g(x))=1/(1-(x/(1-x)))
=1/((1-x-x)/(1-x))
=1/((1-2x)/(1-x))
=(1-x)/(1-2x)
2007-01-27 10:14:16 · answer #5 · answered by Anonymous · 1⤊ 0⤋