Question

Let $f$ and $g$ be the functions defined by $f(x) = \dfrac{x}{{1 + x}}$ and $g(x) = \dfrac{x}{{1 - x}}$. Then $fo{g^{ - 1}}(x) = $ ?
A) $x$
B) $2x$
C) $3x$
D) $4x$

Answer 1

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

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

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