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Question: Let \(f\) and \(g\) be the functions defined by \(f(x) = \dfrac{x}{{1 + x}}\) and \(g(x) = \dfrac{x}...

Let ff and gg be the functions defined by f(x)=x1+xf(x) = \dfrac{x}{{1 + x}} and g(x)=x1−xg(x) = \dfrac{x}{{1 - x}}. Then fog−1(x)=fo{g^{ - 1}}(x) = ?
A) xx
B) 2x2x
C) 3x3x
D) 4x4x

Explanation

Solution

The term fogfog means a composite function of f and gf{\text{ and }}g. And can be denoted as:
fog=f(g(x))fog = f(g(x))
But we have to find the inverse of this composite function its inverse will be written as:
(fog)−1  (x) = (g−1  o f−1) (x){\left( {fog} \right)^{ - 1}}\;\left( x \right){\text{ }} = {\text{ }}({g^{ - 1}}\;o{\text{ }}{f^{ - 1}}){\text{ }}\left( x \right)
That means we will find first the inverse of the two given function and find fog−1(x)fo{g^{ - 1}}(x)

Complete step by step answer:
We are given :f(x)=x1+xf(x) = \dfrac{x}{{1 + x}}
And g(x)=x1−xg(x) = \dfrac{x}{{1 - x}}
We have to find the fog−1(x)fo{g^{ - 1}}(x).
Since: (fog)−1  (x) = (g−1  o f−1) (x){\left( {fog} \right)^{ - 1}}\;\left( x \right){\text{ }} = {\text{ }}({g^{ - 1}}\;o{\text{ }}{f^{ - 1}}){\text{ }}\left( x \right)
We will find the inverses of both the function:
Inverse of Function f(x)f(x):
f(x)=y=x1+xf(x) = y = \dfrac{x}{{1 + x}}
⇒y(x+1)=x\Rightarrow y(x + 1) = x
⇒x=yx+y\Rightarrow x = yx + y
To find inverse we try to express this function in terms of yy instead of xx
⇒x (1 − y) = y\Rightarrow x{\text{ }}\left( {1{\text{ }}-{\text{ }}y} \right){\text{ }} = {\text{ }}y
⇒x = y [1 − y] \Rightarrow x{\text{ }} = \dfrac{{{\text{ }}y}}{{{\text{ }}\left[ {1{\text{ }}-{\text{ }}y} \right]}}{\text{ }}
f−1  (y) = y[1 − y] {f^{ - 1}}\;\left( y \right){\text{ }} = \dfrac{{{\text{ }}y}}{{\left[ {1{\text{ }}-{\text{ }}y} \right]}}{\text{ }}
f−1  (x) = x 1−x{f^{ - 1}}\;\left( x \right){\text{ }} = {\text{ }}\dfrac{{x{\text{ }}}}{{1 - x}}
The inverse of function g(x)g(x)
Let z = g (x) = x[1 − x] z{\text{ }} = {\text{ }}g{\text{ }}\left( x \right){\text{ }} = \dfrac{{{\text{ }}x}}{{\left[ {1{\text{ }}-{\text{ }}x} \right]}}{\text{ }}
⇒z − zx = x\Rightarrow z{\text{ }}-{\text{ }}zx{\text{ }} = {\text{ }}x
⇒x (z + 1) = z\Rightarrow x{\text{ }}\left( {z{\text{ }} + {\text{ }}1} \right){\text{ }} = {\text{ }}z
⇒x = z z + 1\Rightarrow x{\text{ }} = \dfrac{{{\text{ }}z{\text{ }}}}{{z{\text{ }} + {\text{ }}1}}
g−1  (z) = zz + 1 {g^{ - 1}}\;\left( z \right){\text{ }} = \dfrac{{{\text{ }}z}}{{z{\text{ }} + {\text{ }}1}}{\text{ }}
Expressing in the terms of xx,
g−1  (x) = x x +1 {g^{ - 1}}\;\left( x \right){\text{ }} = \dfrac{{{\text{ }}x{\text{ }}}}{{x{\text{ }} + {\text{1}}}}{\text{ }}
Now we will find fog−1(x)fo{g^{ - 1}}(x):
(fog)−1  (x) = g−1  (f−1  (x)){\left( {fog} \right)^{ - 1}}\;\left( x \right){\text{ }} = {\text{ }}{g^{ - 1}}\;({f^{ - 1}}\;\left( x \right))
g−1  (x1−x)=x1 − x1+x1 − x{g^{ - 1}}\;\left( {\dfrac{x}{{1 - x}}} \right) = \dfrac{{\dfrac{x}{{1{\text{ }}-{\text{ }}x}}}}{{1 + \dfrac{x}{{1{\text{ }}-{\text{ }}x}}}}
g−1  (x1−x)=x1 − x1−x+x1 − x{g^{ - 1}}\;\left( {\dfrac{x}{{1 - x}}} \right) = \dfrac{{\dfrac{x}{{1{\text{ }}-{\text{ }}x}}}}{{\dfrac{{1 - x + x}}{{1{\text{ }}-{\text{ }}x}}}}
g−1  (x1−x)=x1 − x11 − x{g^{ - 1}}\;\left( {\dfrac{x}{{1 - x}}} \right) = \dfrac{{\dfrac{x}{{1{\text{ }}-{\text{ }}x}}}}{{\dfrac{1}{{1{\text{ }}-{\text{ }}x}}}}
g−1  (x1−x)=x{g^{ - 1}}\;\left( {\dfrac{x}{{1 - x}}} \right) = x
Hence, fog−1(x)=xfo{g^{ - 1}}(x) = x. So, option (A) is correct.

Note:
The composite functions are not commutative that means that:
fog≠goffog \ne gof,
The order is important and thus if we had to find fog−1fo{g^{ - 1}} the values would have been different.