Question

If $f(x)=2x-6$, then ${{f}^{-1}}(x)$ is:
$6-2x$
A.$\dfrac{1}{2}x-6$
B.$\dfrac{1}{2}x-3$
C.$\dfrac{1}{2}x+3$
D.$\dfrac{1}{2}x+6$

Answer 1

Hint: A function from X to Y is a rule or correspondence that assigns to each element of set Y, one and only one element of set Y. If a function is both one-one and onto then there exists a unique function which is known as the inverse of a given function. Using this methodology, we can solve the given problem.

Complete step by step answer:
Functions can be easily defined with the help of the concept of mapping. Let X and Y be any two non-empty sets. A function from X to Y is a rule or correspondence that assigns to each element of set Y, one and only one element of set Y. Mathematically we write: $f:X\to Y$ where $y=f(x),x\in X\text{ and }y\in Y$. Every element in set X should have one and only one image. Set X is called the domain of the function ‘f’ and set Y is called the codomain of the function ‘f’.
Let $f:X\to Y$ be a function defined by $y=f(x)$ such that f is both one-one and onto. Then there exists a unique function $g:Y\to X$ such that for each $y\in Y$.
$g(y)=x\Leftrightarrow y=f(x).$
The function g is called the inverse of f.
As per our problem, we are given $f(x)=2x-6$. Since, f(x) is one-one and onto, it is invertible.
Also, we know that $f({{f}^{-1}}(x))=x$
$\begin{align}
  & \therefore 2{{f}^{-1}}(x)-6=x \\
 & 2{{f}^{-1}}(x)=x+6 \\
 & {{f}^{-1}}(x)=\dfrac{x+6}{2} \\
 & {{f}^{-1}}(x)=\dfrac{x}{2}+3 \\
\end{align}$
Therefore, option (b) is correct.

Note: The key step for solving this problem is the knowledge of the inverse of a function. First, we must know conclusively that the function is one-one and onto. If a function is not one-one or onto, then its inverse is not possible. Hence, after this we proceed by using $f({{f}^{-1}}(x))=x$.