Question

How do you find the inverse of $f(x) = \dfrac{1}{{2x}}$?

Answer 1

This question is from the topic of inverses of function. In this question we need to find the inverse of function $f(x) = \dfrac{1}{{2x}}$. To find the inverse of a first we need to check that the given function is bijective i.e., one-one and onto function then the inverse of a function exists. A function inverse exists if and only if it is a bijective function.

Complete step by step answer:
Let us try to solve this question in which we are asked to find the inverse of function $f(x) = \dfrac{1}{{2x}}$. Before solving this question we will first recall the definition of one-one and onto function, since we have to prove $f(x) = \dfrac{1}{{2x}}$ to one-one and onto function then only it can have inverse.

One-one function: A function $f:X \to Y$ is defined to be one-one if $\forall \,{x_1},{x_2}\, \in \,X,\,f({x_1}) = \,f({x_2})\,\, \Rightarrow \,\,{x_1} = {x_2}$
Onto function: A function$f:X \to Y$ is defined to be onto if for all $y\,\in Y$ there exists $x\, \in X$ such that $f(x) = y$.
So, let’s find the inverse of function $f(x) = \dfrac{1}{{2x}}$.
Assuming domain and range of function $f(x) = \dfrac{1}{{2x}}$ to be $\Re \\{ 0\\}$.

To Prove: $f(x) = \dfrac{1}{{2x}}$ is one-one function where $f$ is defined on $f:\Re \\{ 0\\} \to \Re \\{ 0\\}$.
Proof: Suppose for every ${x_1},{x_2}$, we have $f({x_1}) = f({x_2})$. We will prove that then ${x_1} = {x_2}$.
f({x_1}) = f({x_2}) \\\ \Rightarrow\dfrac{1}{{2{x_1}}} = \dfrac{1}{{2{x_2}}} \\\ \Rightarrow 2{x_1} = 2{x_2} \\\
After cancellation of $2$ from the both side of equation in above equation, we have
${x_1} = {x_2}$
Since, we have proven ${x_1} = {x_2}$. Hence the function $f(x) = \dfrac{1}{{2x}}$ is a one-one function.

To prove: $f(x) = \dfrac{1}{{2x}}$ is onto function where $f$ is defined on $f: \Re \\{ 0\\} \to \Re \\{ 0\\}$.
Proof: To prove onto function we will find for every $y$ there exists a $x$. Let$y = f(x)$,
$y = f(x) = \dfrac{1}{{2x}}$.......................$(1)$
Now multiplying both of the equation $(1)$ by $\dfrac{x}{y}$, we get
$x = \dfrac{1}{{2y}}$..........................$(2)$
Since we have found $x$ for every $y$. Hence the given function$f(x) = \dfrac{1}{{2x}}$.
Since the function $f(x) = \dfrac{1}{{2x}}$ is both one-one and onto. Hence the inverse of function $f(x) = \dfrac{1}{{2x}}$ exists and its inverse is given by ${f^{ - 1}}(x) = \dfrac{1}{{2x}}$.

Note: Given function $f(x) = \dfrac{1}{{2x}}$ is not defined at $x = 0$ that’s why we have excluded $0$ from its domain. To solve these types of questions we need to know the definitions of one-one and onto function without it we cannot find the inverse of a function.