Question

If $f\left( x \right)=\dfrac{x+1}{x-1}\text{ }\left( x\ne \pm 1 \right)$ then find

a)$\left( fofof \right)\left( x \right)$
b)$\left( fofofof \right)\left( x \right)$

Answer 1

Hint: The O in the expression, represents the word of. $\left( fof \right)\left( x \right)$ is also pronounced as f of f of x. So, it means find $f\left( x \right)$ . Substitute the $f\left( x \right)$ in the place of x to get $f\left( f\left( x \right) \right)$ which nothing but $\left( fof \right)\left( x \right)$ . So, here in the question he asked for 3 and 4 fs. So, do the above process thrice for part (i) 4 times for part (ii). Simply substitute given function into the given function as variable.

Complete step-by-step answer:

Given function in the question is expressed by the f:

$f\left( x \right)=\dfrac{x+1}{x-1}$

So, for part (i) we need $\left( fofof \right)\left( x \right)$ value.

The symbol O represents. So, the above expression can be:

$\left( fofofo \right)\left( x \right)=f\text{ of }f\text{ of }f\text{ of }x=f\left( f\left( f\left( x \right) \right) \right)$

Given we know the function $f\left( x \right)$ , given by the expression:

$f\left( x \right)=\dfrac{x+1}{x-1}$

Now, we will substitute $x=f\left( x \right)=\dfrac{x+1}{x-1}$ in the equation above to find $fof\left( x \right)$.

By substituting the above condition, we can write it as:

$f\left( \dfrac{x+1}{x-1} \right)=\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}\text{ }$

By taking least common multiple and cancelling $\left( x-1 \right)$ , we get

$f\left( \dfrac{x+1}{x-1} \right)=\dfrac{x+1+x-1}{x+1-x+1}=\dfrac{2x}{2}$

By simplifying the above term, we get the equation:

$\left( fof \right)\left( x \right)=x$

By substituting x into again function we get it as:

$f\left( \left( fof \right)\left( x \right) \right)=\dfrac{x+1}{x-1}$

By above we can say $fofof\left( x \right)$ value to be given by:

$\left( fofof \right)\left( x \right)=\dfrac{x+1}{x-1}$

By substituting this as x into the function back, we get

$f\left( \left( fofof \right)\left( x \right) \right)=\dfrac{\dfrac{x+1}{x-1}+1}{\dfrac{x+1}{x-1}-1}=\dfrac{x+1+x-1}{x+1-x+1}$

By simplifying this equation, we get $\left( fofofof \right)\left( x \right)$ value to x:

$\left( fofofof \right)\left( x \right)=x$

By the above 2 equations we found values of $\left( fofof \right)\left( x \right)$ , $\left( fofofof \right)\left( x \right)$ as asked.

Hence, provided the required expressions.

Note: Observe that all the even times function is given by the x and odd times function given by $\dfrac{x+1}{x-1}$ . So, by this we can find any number of times the function.