Question

Consider $f:R\to R$ given by f(x)=4x+3. Show that f is invertible. Find the inverse of f.

Answer 1

Here, $f:R\to R$ is given by f(x) =4x+3
Let x, y in R such that
$f\left(x\right)=f\left(y\right)\Rightarrow 4x+3=4y+3$
$\Rightarrow 4x=4y\Rightarrow x=y$
Therefore, f is a one-one function.
Let y =4x+3
$\Rightarrow$ There exist, $x=\left(\frac{y-3}{4}\right)\in R,\forall y\in R$
Therefore, for any y in R, there exist $x=\frac{y-3}{4}\in R$ such that
$f\left(x\right)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y$
Therefore, f is onto function.

Thus, f is one-one and onto and therefore, $f^{-1}$ exists.
Let us define $g:R\to R$ by $g\left(x\right)=\frac{x-3}{4}$
Now, $\left(gof\right)\left(x\right)=g\left(f\right(x\left)\right)=g(4x+3)=\frac{(4x+3)-3}{4}=x$
$\left(fog\right)\left(y\right)=f\left(g\right(y\left)\right)=f\left(\frac{y-3}{4}\right)=4\left(\frac{y-3}{4}\right)+3=y-3+3=y$
Therefore, gof $=fog=I_R$
Hence, f is invertible and the inverse of f is given by $f^{-1}\left(y\right)=g\left(y\right)=\frac{y-3}{4}$