Let f : Q → Q : f(x) = 3x —4. Show that f is invertible and find $f^{-1}$.

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Let f : Q → Q : f(x) = 3x —4. Show that f is invertible and find $f^{-1}$ .

Let f : Q → Q : f(x) = 3x —4. Show that f is invertible and find $f^{-1}$ .

Solution:

$\mathrm{f}(\mathrm{x})=3 \mathrm{x}-4 \quad$ (as given)
$\mathrm{f}(\mathrm{x})$ is invertible if $\mathrm{f}(\mathrm{x})$ is a bijection (i.e one-one onto function)
One-One function
Suppose $\mathrm{p}, \mathrm{q}$ be two arbitrary elements in $\mathrm{R}$
Therefore, $f(p)=f(q)$
$\begin{array}{l} \Rightarrow 3 \mathrm{p}-4=3 \mathrm{q}-4 \\ \Rightarrow 3 \mathrm{p}=3 \mathrm{q} \\ \Rightarrow \mathrm{p}=\mathrm{q} \end{array}$
When $f(p)=f(q), p=q$
Therefore, $\mathrm{f}(\mathrm{x})$ is one-one function.
Onto function
Suppose $\mathrm{v}$ be an arbitrary element of $\mathrm{R}$ (Co-domain)
Then, $f(x)=v$
$3 x-4=v$
$\Rightarrow 3 \mathrm{x}=\mathrm{v}+4$ $\Rightarrow \mathrm{x}=\frac{v+4}{3}$ Since $\mathrm{v} \in \mathrm{Q}$
$\begin{array}{l} \Rightarrow \frac{v+4}{3} \in Q \\ f(x)=f\left(\frac{v+4}{3}\right) \end{array}$
$\begin{aligned} &=3\left(\frac{v+4}{3}\right)-4 \\ &=\mathrm{v}+4-4=\mathrm{v} \\ \Rightarrow \mathrm{f}(\mathrm{x}) &=\mathrm{v} \end{aligned}$
Therefore this shows that every element in the co-domain has its pre-image in domain.
Hence, $f(x)$ is onto function.
As a result, $f(x)$ is invertible.
Now we need to find $\mathrm{f}_{-1}$ ,
Suppose $\mathrm{f}(\mathrm{x})=\mathrm{y}$
$y=3 x-4$
Now, replace all $x$ with y and all y with $x$ .
Now, solve for y
$\begin{array}{l} \Rightarrow x+4=3 y \\ \Rightarrow y=\frac{x+4}{3} \end{array}$
Now replace $\mathrm{y}$ with $\mathrm{f}_{-1}(\mathrm{x})$
As a result, $\mathrm{f}_{-1}(\mathrm{x})=\frac{x+4}{3}$