Question

Let $f:N\to Y$ 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 and its inverse is

• A. $g(y)=\frac{y-3}{4}$
• B. $g(y)=\frac{3y+4}{3}$
• C. $g(y)=4+\frac{y+3}{4}$
• D. $g(y)=\frac{y+3}{4}$

Answer 1

Answer: (A)

Let $f(x_1)=f(x_2),x_1,x_2\in N$
$\Rightarrow 4x_1+3=4x_2+3$
$\Rightarrow x_1=x_2$
Thus $f(x_1)=f(x_2)$
$\Rightarrow x_1=x_2$.
Hence the function is one-one.
Let $y\in Y$ be a number of the form
$y=4k+3$, for some $k\in N$, then
$y=f(x)$
$\Rightarrow 4k+3=4x+3$
$\Rightarrow x=k\in N$
Thus corresponding to any $y\in Y$ we have $x\in N$.
The function then is onto.
The function, being both one-one and onto is invertible.
$y=4x+3$
$\Rightarrow x=\frac{y-3}{4}$
$\therefore f^{-1}(x)=\frac{x-3}{4}$
or $g(y)=\frac{y-3}{4}$
is the inverse of the function.