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.