Question
Let $$f:N \to Y$$ be a function defined as $$f\left( x \right) = 4x + 3$$ where $$Y = \left\{ {y \in N:y = 4x + 3\,{\text{for}}\,{\text{some}}\,x \in N} \right\}.$$
Show that $$f$$ is invertible and its inverse is
A.
$$g\left( y \right) = \frac{{3y + 4}}{3}$$
B.
$$g\left( y \right) = 4 + \frac{{y + 3}}{4}$$
C.
$$g\left( y \right) = \frac{{y + 3}}{4}$$
D.
$$g\left( y \right) = \frac{{y - 3}}{4}$$
Answer :
$$g\left( y \right) = \frac{{y - 3}}{4}$$
Solution :
Clearly $$f$$ is one one and onto, so invertible
$$\eqalign{
& {\text{Also}}\,f\left( x \right) = 4x + 3 = y \Rightarrow x = \frac{{y - 3}}{4} \cr
& \therefore g\left( y \right) = \frac{{y - 3}}{4} \cr} $$