Question

If $f\left( x \right) = \dfrac{x}{{x - 1}}$, $x \ne 1$. Then what is the value of ${\left( {fofof....f} \right)_{17 times}}\left( x \right)$?
A. $\dfrac{x}{{x - 1}}$
B. $x$
C. ${\left( {\dfrac{x}{{x - 1}}} \right)^{17}}$
D. $\dfrac{{17x}}{{x - 1}}$

Answer 1

Hint: Here use the definition of a composite function and find the values. Check the composition for more functions and verify the similarities to reach the required answer.

Formula Used:
A composite function is a function created when one function is used as the input of another function.
$\left( {fof} \right)\left( x \right) = f\left( {f\left( x \right)} \right)$

Complete step by step solution:
 The given function is if $f\left( x \right) = \dfrac{x}{{x - 1}}$, $x \ne 1$.
Let’s calculate the composite function of two functions.
$\left( {fof} \right)\left( x \right) = f\left( {f\left( x \right)} \right)$
Substitute the value of the given function.
$\left( {fof} \right)\left( x \right) = f\left( {\dfrac{x}{{x - 1}}} \right)$
$ \Rightarrow $$\left( {fof} \right)\left( x \right) = \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{x}{{x - 1}} - 1}}$
Simplify the above equation.
$\left( {fof} \right)\left( x \right) = \dfrac{{\dfrac{x}{{x - 1}}}}{{\dfrac{{x - x + 1}}{{x - 1}}}}$
$ \Rightarrow $$\left( {fof} \right)\left( x \right) = \dfrac{x}{1}$
$ \Rightarrow $$\left( {fof} \right)\left( x \right) = x$ $.....\left( 1 \right)$

Now calculate the composite function of 3 functions.
$\left( {fofof} \right)\left( x \right) = f\left( {fof\left( x \right)} \right)$
Substitute the equation $\left( 1 \right)$ in the above function.
$\left( {fofof} \right)\left( x \right) = f\left( x \right)$
$ \Rightarrow $$\left( {fofof} \right)\left( x \right) = \dfrac{x}{{x - 1}}$ $.....\left( 2 \right)$

Now calculate the composite function of 4 functions.
$\left( {fofofof} \right)\left( x \right) = f\left( {fofof\left( x \right)} \right)$
Substitute the equation $\left( 2 \right)$ in the above function.
$\left( {fofofof} \right)\left( x \right) = f\left( {\dfrac{x}{{x - 1}}} \right)$
Using the previous steps of equation $\left( 1 \right)$, we get
$\left( {fofofof} \right)\left( x \right) = x$

Clearly, we observe that
For the odd number of functions, the value of composite functions is $\dfrac{x}{{x - 1}}$.
For the even number of functions, the value of composite functions is $x$.
Since $17$ is an odd number.
So, ${\left( {fofof....f} \right)_{17 times}}\left( x \right) = \dfrac{x}{{x - 1}}$

Option ‘A’ is correct

Note: The composition of functions is a process of combining two functions. One function is performed first and its result is substituted as the input value of the second function.
The composite function of the two functions $f\left( x \right)$ and $g\left( x \right)$ is denoted as $f\left( {g\left( x \right)} \right)$ or $\left( {fog} \right)\left( x \right)$.