Show that the function f : R → R : f(x) = 2x + 3 is invertible and find $f^{-1}$.

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Show that the function f : R → R : f(x) = 2x + 3 is invertible and find $f^{-1}$ .

Show that the function f : R → R : f(x) = 2x + 3 is invertible and find $f^{-1}$ .

Solution:

$\mathrm{f}(\mathrm{x})=2 \mathrm{x}+3$ (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, $f(p)=f(q)$
$\begin{array}{l} \Rightarrow 2 p+3=2 q+3 \\ \Rightarrow 2 p=2 q \\ \Rightarrow p=q \end{array}$
When $f(p)=f(q), p=q$
Therefore, $\mathrm{f}(\mathrm{x})$ is one-one function.
Onto function
Suppose $v$ be an arbitrary element of $R$ (Co-domain)
Therefore, $f(x)=v$
$\begin{array}{l} 2 \mathrm{x}+3=\mathrm{v} \\ \Rightarrow 2 \mathrm{x}=\mathrm{v}-3 \\ \Rightarrow \mathrm{x}=\frac{v-3}{2} \end{array}$
As $\mathrm{v} \in \mathrm{R}$
$\Rightarrow \frac{v-3}{2} \in \mathrm{R}$
$\begin{array}{l} \begin{aligned} f(x) &=f\left(\frac{v-3}{2}\right) \\ &=2\left(\frac{v-3}{2}\right)+3 \\ &=v-3+3=v \\ \Rightarrow f & \end{aligned} \\ \Rightarrow f(x)=v \end{array}$
It shows that every element in the co-domain has its pre-image in domain.
Therefore, $f(x)$ is onto function.
Thus, $f(x)$ is invertible.
Now we need to find $\mathrm{f}-1$ ,
Suppose $\mathrm{f}(\mathrm{x})=\mathrm{y}$
$y=2 x+3$
Now, replace all $x$ with y and all y with $x$ .
Now, solve for $\mathrm{y}$
$\begin{array}{l} \Rightarrow x-3=2 y \\ \Rightarrow y=\frac{x-3}{2} \end{array}$
Now replace y with $\mathrm{f}_{-1}(\mathrm{x})$
As a result, $\mathrm{f}_{-1}(\mathrm{x})=\frac{x-3}{2}$