Question

How do you find the inverse of $f(x)=1-\left( \dfrac{1}{x} \right)$?

Answer 1

An inverse function is defined as a function, which can reverse into another function, or if a function takes $x$ to $y$ then the inverse function takes $y$ to $x$. If there is a function $f(x)$ having an inverse function $g(x)$. Then the domain of the function $f(x)$ is the range of the function $g(x)$ and the domain of $g(x)$ is the range of the function $f(x)$. The inverse of a function is represented as ${{f}^{-1}}(x)$, one shouldn’t confuse the -1 with the exponent here. Once we know that the inverse function exists, it can be found by following the steps below
1. First, replace $f(x)$ with $y$.
2. Replace every $x$ with a $y$ and replace every $y$ with an $x$ .
3. Solve the equation from Step 2 for $y$.
4. Replace $y$ with ${{f}^{-1}}(x)$.

Complete step by step answer:
The given function is $f(x)=1-\left( \dfrac{1}{x} \right)$, we have to find its inverse. To find the inverse of a function, we have to follow a set of certain steps as follow,
We will now these steps accordingly, in the given function $f(x)=1-\left( \dfrac{1}{x} \right)$, replacing $f(x)$ with $y$ we get, $y=1-\left( \dfrac{1}{x} \right)$
Replacing every $x$ with a $y$ and every $y$ with an $x$, we get $x=1-\left( \dfrac{1}{y} \right)$
Subtracting $1$ from both sides of the above equation we get,
$\Rightarrow x-1=1-\left( \dfrac{1}{y} \right)-1$
$\Rightarrow x-1=-\left( \dfrac{1}{y} \right)$
Multiplying $\dfrac{y}{(x-1)}$ to both sides of the equation, we get
$\Rightarrow \left( x-1 \right)\dfrac{y}{(x-1)}=-\left( \dfrac{1}{y} \right)\dfrac{y}{(x-1)}$
$\therefore y=\dfrac{-1}{x-1}$

Finally replacing $y$ with ${{f}^{-1}}(x)$ we get, ${{f}^{-1}}(x)=\dfrac{-1}{x-1}$ is the inverse of the given function.

Note: We can check whether our answer is correct or not by checking the domain and range of function.
For function $f(x)=1-\left( \dfrac{1}{x} \right)$, Domain $\in (-\infty ,0)\cup (0,\infty )$ and Range $\in (-\infty ,1)\cup (1,\infty )$
And for function $g(x)=\dfrac{-1}{x-1}$, Domain $\in (-\infty ,1)\cup (1,\infty )$ and Range $\in (-\infty ,0)\cup (0,\infty )$
So as the Domain of one function is the range of the other and the range of one is the domain of the other, our answer is correct.