Question

Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f⁻¹

Answer 1

Injectivity of f:
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)
$\Rightarrow$3x + 5 =3y + 5
$\Rightarrow$3x = 3y
$\Rightarrow$x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y

$$\begin{gathered} \Rightarrow 3x+5=y \\ \Rightarrow 3x=y-5 \\ \Rightarrow x=\frac{y-5}{3}\in Q\left(\text{domain}\right) \end{gathered}$$

$\Rightarrow$f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow x=f\left(y\right) \\ \Rightarrow x=3y+5 \\ \Rightarrow x-5=3y \\ \Rightarrow y=\frac{x-5}{3} \\ \text{So,}{f}^{-1}\left(x\right)=\frac{x-5}{3}[\text{from}\left(1\right)] \end{gathered}$$