If f(x) = f(x+1) , then f(f(f(y+2))) is =
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Answered by
0
Answer:
If f(x)=
x+1
x−1
, then
f[f{f(x)}]=f[f(
x+1
x−1
)]
=f
⎣
⎢
⎢
⎡
x+1
x−1
+1
x+1
x−1
−1
⎦
⎥
⎥
⎤
=f
⎣
⎢
⎢
⎡
x+1
x−1+x+1
x+1
x−1−x−1
⎦
⎥
⎥
⎤
=f[−
2x
2
]=f(−
x
1
)
Now, f(−
x
1
)=
−
x
1
+1
−
x
1
−1
=
x
−1+x
x
−1−x
=−
x−1
x+1
=−
f(x)
1
∴f[f{f(x)}]=−
f(x)
1
Answered by
1
Answer:
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