Consider $f: R \rightarrow R$ given by $f(x)=4 x+3$. Show that $f$ is invertible. Find the inverse of $f$.
Injectivity of $f$ :
Let $x$ and $y$ be two elements of domain $(R)$, such that
$f(x)=f(y)$
$\Rightarrow 4 x+3=4 y+3$
$\Rightarrow 4 x=4 y$
$\Rightarrow x=y$
So, $f$ is one-one.
Surjectivity of $f$ :
Let $y$ be in the co-domain $(R)$, such that $f(x)=y$.
$\Rightarrow 4 x+3=y$
$\Rightarrow 4 x=y-3$
$\Rightarrow x=\frac{y-3}{4} \in R($ Domain $)$
$\Rightarrow f$ is onto.
So, $f$ is a bijection and, hence, is invertible.
Finding $f^{-1}$ :
Let $f^{-1}(x)=y$ $\ldots(1)$
$\Rightarrow x=f(y)$
$\Rightarrow x=4 y+3$
$\Rightarrow x-3=4 y$
$\Rightarrow y=\frac{x-3}{4}$
So, $f^{-1}(x)=\frac{x-3}{4} \quad[$ from (1) $]$
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