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Problem
Problem
The inverse of the function
f
(
x
)
=
e
x
−
e
−
x
e
x
+
e
−
x
+
2
f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + 2
f
(
x
)
=
e
x
+
e
−
x
e
x
−
e
−
x
+
2
is given by (a)
log
e
(
x
−
2
x
−
1
)
1
/
2
\log_e \left(\frac{x-2}{x-1}\right)^{1/2}
lo
g
e
(
x
−
1
x
−
2
)
1/2
(b)
log
e
(
x
−
1
3
−
x
)
1
/
2
\log_e \left(\frac{x-1}{3-x}\right)^{1/2}
lo
g
e
(
3
−
x
x
−
1
)
1/2
(c)
log
e
(
x
2
−
x
)
1
/
2
\log_e \left(\frac{x}{2-x}\right)^{1/2}
lo
g
e
(
2
−
x
x
)
1/2
(d)
log
e
(
x
−
1
x
+
1
)
−
2
\log_e \left(\frac{x-1}{x+1}\right)^{-2}
lo
g
e
(
x
+
1
x
−
1
)
−
2
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