2006-09-06 05:25:39 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
well am trying to explain for you step by step .
Step 1 ;
if f(x) = 1-x & g9x) = 1/x ; so
A = g(f(x)) = 1/ (1 - x)
Step 2;
B = f(A) = 1 - ( 1/ (1 - x) )
Step 3;
C = f(B) = 1 - ( 1 - ( 1/ (1 - x) ) )
. Now we have to simplfy this ;
if
f(f(g(f(x)))) = 1 - ( 1 - ( 1/ (1 - x) ) ) =
1 - 1 - ( 1/(1-x)) = 0 - ( 1/(1-x))
so the result is ;
-1/ (1-x) Or 1/(1-x)
Good Luck & Good Question.
2006-09-06 06:04:04 · answer #1 · answered by sweetie 5 · 1⤊ 0⤋
1-(1-1/(1-x)) = -1/(1-x)
2006-09-06 12:32:07 · answer #2 · answered by Morgy 4 · 0⤊ 0⤋
Working inside out....
f(x) = 1-x g(f(x)) = 1-1/x f (1-1/x) = 1-(1/1-x) f(1-(1/1-x)=
1-(1/1-1-x) = 1-1/-x = ( -x - 1) / -x or (x+1) / x
i got (x+1) / x
2006-09-06 12:34:01 · answer #3 · answered by Brian D 5 · 0⤊ 0⤋
Let's start with the inner stuff
f(x) = 1-x
g(f(x)) = 1 / (1-x)
f(g(f(x))) = 1 - (1 / (1 - x))
f(f(g(f(x)))) = 1 - (1 - (1 / (1 - x))) = 1 - 1 + (1 / 1 - x) = 1 / (1-x)
It is not correct.
2006-09-06 12:32:25 · answer #4 · answered by Anonymous · 2⤊ 0⤋
g(f(x)) =
g(1 - x) =
1/(1 - x) =
(1/(-x + 1)) =
(1/(-(x - 1))) =
(-1/(x - 1))
f(g(f(x))) =
f(-1/(x - 1)) =
1 - (-1/(x - 1)) =
1 + (1/(x - 1)) =
((x - 1) + 1)/(x - 1) =
(x - 1 + 1)/(x - 1) =
(x/(x - 1))
f(f(g(f(x)))) =
f(x/(x - 1)) =
1 - (x/(x - 1)) =
((x - 1) - x)/(x - 1) =
(x - 1 - x)/(x - 1) =
(-1/(x - 1))
So you are incorrect.
f(f(g(f(x)))) = (-1/(x - 1)) or (1/(1 - x))
2006-09-06 13:02:16 · answer #5 · answered by Sherman81 6 · 0⤊ 1⤋
the answer is 1 / (1-x)
g(f(x)) = 1 / (1-x)
f(g(f(x))) = 1 - { 1 / (1-x)}
f(f(g(f(x)))) = 1 - [1 - { 1 / (1-x)}] = 1 / (1-x)
2006-09-06 12:35:44 · answer #6 · answered by sandiego_roleplay 1 · 0⤊ 0⤋
Your answer is different. I do not know how you did it
First find g{f(x)}=1/(1-x)
Then find f[g{f(x)}] = 1/1-(1-x)=1/x
Now find f{f[g{f(x)}] } = 1/(1-x)
2006-09-06 12:37:13 · answer #7 · answered by Amar Soni 7 · 0⤊ 0⤋
I think it is (1-2x)/(1-x)
2006-09-06 12:32:02 · answer #8 · answered by SP 1 · 0⤊ 0⤋
Ask your teacher
2006-09-06 12:27:54 · answer #9 · answered by Anonymous · 0⤊ 0⤋