Question

Given $f\left( x \right)=\dfrac{1}{x-2}$ , how do you find $\left( f\left( f\left( \dfrac{1}{2} \right) \right) \right)$ ?
(a) Putting the value step by step
(b) Checking the continuity
(c) Checking the differentiability
(d) None of these

Answer 1

Hint: We are trying to find the value of $\left( f\left( f\left( \dfrac{1}{2} \right) \right) \right)$, where we are given $f\left( x \right)=\dfrac{1}{x-2}$. To find the value of this we will start with find the value $f\left( \dfrac{1}{2} \right)$.Once, we get that value, we can put the value of x on the place of $f\left( x \right)=\dfrac{1}{x-2}$in the function. Then simplifying with the numerator and denominator we will get the value we need.

Complete step by step solution:
According to the question, we are given the function, $f\left( x \right)=\dfrac{1}{x-2}$.
We are to find the value of $\left( f\left( f\left( \dfrac{1}{2} \right) \right) \right)$.
We will start with finding the value of $f\left( \dfrac{1}{2} \right)$ ,
So, putting the value as $x=\dfrac{1}{2}$, in $f\left( x \right)=\dfrac{1}{x-2}$, we get,
$f\left( \dfrac{1}{2} \right)=\dfrac{1}{\dfrac{1}{2}-2}$
Simplifying the denominator,
$\Rightarrow f\left( \dfrac{1}{2} \right)=\dfrac{1}{-\dfrac{3}{2}}$
After more simplifying,
$\Rightarrow f\left( \dfrac{1}{2} \right)=-\dfrac{2}{3}$
Now, we will try to find out the value of $\left( f\left( f\left( \dfrac{1}{2} \right) \right) \right)$,so we have to put the value as $x=-\dfrac{2}{3}$, in $f\left( x \right)=\dfrac{1}{x-2}$,
Again, we are getting,
$f\left( f\left( \dfrac{1}{2} \right) \right)=\dfrac{1}{-\dfrac{2}{3}-2}$
Simplifying the denominator,
$\Rightarrow f\left( f\left( \dfrac{1}{2} \right) \right)=\dfrac{1}{-\dfrac{8}{3}}$
After more simplifying,
$\Rightarrow f\left( f\left( \dfrac{1}{2} \right) \right)=-\dfrac{3}{8}$
Thus, we get the value of $\left( f\left( f\left( \dfrac{1}{2} \right) \right) \right)$as $-\dfrac{3}{8}$.

So, the correct answer is “Option a”.

Note: One of the more important ideas about functions is that of the domain and range of a function. In simplest terms the domain of a function is the set of all values that can be plugged into a function and have the function exist and have a real number for a value. So, for the domain we need to avoid division by zero, square roots of negative numbers, logarithms of zero and logarithms of negative numbers (if not familiar with logarithms we’ll take a look at them a little later), etc. The range of a function is simply the set of all possible values that a function can take.