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Question: If \(f(x)=2x-6\), then \({{f}^{-1}}(x)\) is: \(6-2x\) A.\(\dfrac{1}{2}x-6\) B.\(\dfrac{1}{2}...

If f(x)=2x−6f(x)=2x-6, then f−1(x){{f}^{-1}}(x) is:
6−2x6-2x
A.12x−6\dfrac{1}{2}x-6
B.12x−3\dfrac{1}{2}x-3
C.12x+3\dfrac{1}{2}x+3
D.12x+6\dfrac{1}{2}x+6

Explanation

Solution

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→Yf:X\to Y where y=f(x),x∈X and y∈Yy=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→Yf:X\to Y be a function defined by y=f(x)y=f(x) such that f is both one-one and onto. Then there exists a unique function g:Y→Xg:Y\to X such that for each y∈Yy\in Y.
g(y)=x⇔y=f(x).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−6f(x)=2x-6. Since, f(x) is one-one and onto, it is invertible.
Also, we know that f(f−1(x))=xf({{f}^{-1}}(x))=x
∴2f−1(x)−6=x 2f−1(x)=x+6 f−1(x)=x+62 f−1(x)=x2+3 \begin{aligned} & \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{aligned}
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))=xf({{f}^{-1}}(x))=x.