Let `f: N->Y` be a function defined as `f(x)=4x+3` , where `Y={y in N : y=4x+3` for some `x in N}` . Show that `f` is invertible. Find its inverse.

Options

A. `g(y)=(y+3)/(4)`
B. `g(y)=(y-3)/(4)`
C. `g(y)=(3y+4)/(3)`
D. `g(y)=4+(y+3)/(4)`

Answer

B

Explanation

For any `x, y in N`
`f(x)=f(y) Rightarrow 4x+3=4y+3 Rightarrow x=y`
`therefore` f is one-one
Clearly, `Y=(y in N: y=4x+3"for some "x in N)="Range (f)"`
`therefore` f:N-Y is onto.
Thus, `f:N to Y` is a bijection and hence invertible.
Let g be the inverse of f. Then.
`fog (y)=y"for all "y inY`
`Rightarrow f(g(y))=yRightarrow 4g(y)+3 =yRightarrow g(y)=(y-3)/(4)`