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Mathematics
Mathematics
Problem
Let
f
:
(
0
,
1
)
→
R
f: (0, 1) \to R
f
:
(
0
,
1
)
→
R
be defined by
f
(
x
)
=
b
−
x
1
−
b
x
f(x) = \frac{b-x}{1-bx}
f
(
x
)
=
1
−
b
x
b
−
x
, where
b
b
b
is a constant such that
0
<
b
<
1
0 < b < 1
0
<
b
<
1
. Then, a.
f
f
f
is not invertible on
(
0
,
1
)
(0, 1)
(
0
,
1
)
b.
f
≠
f
−
1
f \neq f^{-1}
f
=
f
−
1
on
(
0
,
1
)
(0, 1)
(
0
,
1
)
and
f
′
(
b
)
=
1
f
′
(
0
)
f'(b) = \frac{1}{f'(0)}
f
′
(
b
)
=
f
′
(
0
)
1
c.
f
=
f
−
1
f = f^{-1}
f
=
f
−
1
on
(
0
,
1
)
(0, 1)
(
0
,
1
)
and
f
′
(
b
)
=
1
f
′
(
0
)
f'(b) = \frac{1}{f'(0)}
f
′
(
b
)
=
f
′
(
0
)
1
d.
f
−
1
f^{-1}
f
−
1
is differentiable on
(
0
,
1
)
(0, 1)
(
0
,
1
)
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