Question

If $f(x) = \dfrac{1}{{1 - x}}$ , show that $f[f\{ f(x)\} ] = x$.

Answer 1

Hint:As we know that the above question consists of a functional equation.A functional equation is any equation in which the unknown represents the function. We know that this type of function assigns exactly one output to each specified type. It is common to name the functions $f(x)$ or $g(x)$. These functional equations have a common technique for solving the value of $f(x)$. A function $f(x)$ is known as a continuous function.

Complete step by step answer:
As per the given question we have $f(x) = \dfrac{1}{{1 - x}}$ and we have to show that Left hand side i.e. $f[f\{ f(x)\} ]$ is equal to the Right hand side i.e. $x$. So we take the LHS, $f[f\{ f(x)\} ]$ and the solve it by replacing
$x = \dfrac{1}{{1 - x}}$.
So we can write by first replacing $x$ :
$f\left( {f\left( {\dfrac{1}{{1 - x}}} \right)} \right)$
Now we will again replace the $x$ in the denominator with $\dfrac{1}{{1 - x}}$. It gives us:
$f\left( {\dfrac{1}{{1 - \left( {\dfrac{1}{{1 - x}}} \right)}}} \right)$

We will simplify this first by taking the LCM of the denominator, it can be written as
$f\left( {\dfrac{1}{{\dfrac{{1 - x - 1}}{{1 - x}}}}} \right)$
We can take the denominator of the fraction in the denominator above i.e.
$f\left( {\dfrac{{1 - x}}{{1 - x - 1}}} \right) \\
\Rightarrow f\left( {\dfrac{{1 - x}}{{ - x}}} \right)$
We can interchange the sign of the numerator and denominator without changing its meaning i.e. $f\left( {\dfrac{{x - 1}}{x}} \right)$ .

So here we will again replace the $x$ with $\dfrac{1}{{1 - x}}$. We can write the expression as,
$\dfrac{{\left( {\dfrac{1}{{1 - x}} - 1} \right)}}{{\dfrac{1}{{1 - x}}}}$ .
We will now solve this , we will take the LCM in the numerator part i.e.
$\dfrac{{\left( {\dfrac{{1 - 1 + x}}{{1 - x}}} \right)}}{{\dfrac{1}{{1 - x}}}}$
On further solving it can be written as:
$\dfrac{{\left( {\dfrac{x}{{1 - x}}} \right)}}{{\dfrac{1}{{1 - x}}}}$
We can write this expression as
$\dfrac{x}{{1 - x}} \div \dfrac{1}{{1 - x}} \Rightarrow \dfrac{x}{{1 - x}} \times \dfrac{{1 - x}}{1}$.
Therefore on multiplication it gives us the value $x$.Hence it is proved that $f[f\{ f(x)\} ] = x$.

Note:Before solving this type of question we should consider the function equation, their formulas and method to solve it. We should note that in the above question the given function$f(x) = \dfrac{1}{{1 - x}}$ , $x \ne 1$. Because if we put the value of $x = 1$ , we get an undefined value i.e. $\dfrac{1}{{1 - 1}} = \dfrac{1}{0}$ and we know that it is undefined. We should also have the knowledge of the algebraic identities as they are very useful in calculation of this kind of problem. Like we have used above that, if we have $\dfrac{a}{{\dfrac{b}{{\dfrac{m}{n}}}}}$ , then we can write this as $\dfrac{a}{b} \times \dfrac{n}{m}$. We should note that a function is a relation in which each input has only one output.