Problem 7: A function
f
:
[
1,
∞
)
→
[
1,
∞
)
f
:
[
1,
∞
)
→
[
1,
∞
)
is given by :
f
x
=
2
x
x
−
1
f
x
=
2
x
x
−
1
Find
f

1
x
f

1
x
.
Solution :
Statement of the problem : In order to determine the nature of given function to be a bijection, we need to check whether the function is both oneone and onto?
To check for oneone function, we determine the derivative of the function as :
f
x
=
2
x
x
−
1
f
x
=
2
x
x
−
1
Taking logarithm on both sides,
log
e
f
x
=
x
x
−
1
log
e
2
log
e
f
x
=
x
x
−
1
log
e
2
Taking derivative on either side of the equation, we have :
f
′
x
f
x
=
2
x
−
1
log
e
2
⇒
f
′
x
=
f
x
2
x
−
1
log
e
2
f
′
x
f
x
=
2
x
−
1
log
e
2
⇒
f
′
x
=
f
x
2
x
−
1
log
e
2
Now, “x” lie in the interval [1,∞ ). Hence, f(x) is a positive number. Also,
log
e
2
log
e
2
is a positive number. Therefore,
f
′
x
=
positive number
X
2
x
−
1
f
′
x
=
positive number
X
2
x
−
1
The root of the expression of f’(x), when equated to zero, is “0.5”. The derivative is positive in the interval between “1/2” and infinity. However, the domain of function begins at x = 1. We, therefore, conclude that the function is increasing in the interval given by [1,∞). The least value of the function is given as :
⇒
f
1
=
2
x
x
−
1
=
2
0
=
1
⇒
f
1
=
2
x
x
−
1
=
2
0
=
1
We interpret the function, when “x” tends to become positive infinity (this is equivalent of taking limit).
lim
x
→
∞
f
x
=
lim
x
→
∞
2
x
x
−
1
=
2
∞
X
∞
−
1
=
2
∞
X
∞
=
2
∞
=
∞
lim
x
→
∞
f
x
=
lim
x
→
∞
2
x
x
−
1
=
2
∞
X
∞
−
1
=
2
∞
X
∞
=
2
∞
=
∞
Thus, range of the given function is [1, ∞). Clearly,
Range = Codomain = [1, ∞)
Therefore, function is onto and hence bijection. It means that function is invertible. Now, we solve the function for “x” for finding inverse function. Taking log on the base of “2” on either side of the equation,
⇒
log
2
y
=
log
2
x
x
−
1
=
x
x
−
1
=
x
2
−
x
⇒
log
2
y
=
log
2
x
x
−
1
=
x
x
−
1
=
x
2
−
x
⇒
x
2
−
x
−
log
2
y
=
0
⇒
x
2
−
x
−
log
2
y
=
0
⇒
x
=
1
±
1
+
4
X
1
X
log
2
y
2
=
1
±
1
+
4
log
2
y
2
⇒
x
=
1
±
1
+
4
X
1
X
log
2
y
2
=
1
±
1
+
4
log
2
y
2
But, we know that domain of the function is x >= 1. Also,
1
+
4
log
2
y
≥
1
1
+
4
log
2
y
≥
1
. It means that only positive sign in the expression is valid.
⇒
x
=
1
+
1
+
4
log
2
y
2
⇒
x
=
1
+
1
+
4
log
2
y
2
Substituting “x” by “
f

1
x
f

1
x
” and “y” by “x”, we have the rule of inverse function as :
⇒
f

1
x
=
1
+
1
+
4
log
2
x
2
⇒
f

1
x
=
1
+
1
+
4
log
2
x
2