Question

Find the value of \[{f^{ - 1}}\left( x \right)\] if let \[f:R \to R\] be defined by \[f\left( x \right) = 2x + 6\] which is bijective mapping.
A) \[\dfrac{x}{2} - 3\]
B) \[2x + 6\]
C) \[x - 3\]
D) \[6x + 2\]

Answer 1

Hint: we are given that \[f:R \to R\] is bijective mapping. First find x in terms of y by \[f\left( x \right)\]= y. After this you will get \[{f^{ - 1}}\left( y \right)\], but we are asked to find \[{f^{ - 1}}\left( x \right)\] which is found by simply replacing y by x in \[{f^{ - 1}}\left( y \right)\].

Complete step-by-step answer:
Let \[y = f\left( x \right)\]…………..(1)
\[ \Rightarrow \]2x = y – 6
\[ \Rightarrow \]\[x = \dfrac{y}{2} - 3\]= \[{f^{ - 1}}\left( y \right)\] (\[\because \] equation (1) can be written in form of x= \[{f^{ - 1}}\left( y \right)\])
Now replace y with x to find out \[{f^{ - 1}}\left( x \right)\]
Hence, \[{f^{ - 1}}\left( x \right)\]=\[\dfrac{x}{2} - 3\].
∴ The correct option is ‘A’.

Note: Bijective function, one-to-one correspondence, or invertible function, is a function between the elements of two sets, where each element of one set is paired with exactly one element of the other set, and each element of the other set is paired with exactly one element of the first set.