Question

Show that the function $f:R\to R$ defined by $f\left(x\right)=\frac{3x-1}{2},x\in R$ is one-one and onto functions. Also, find the inverse of the function f.

Answer 1

Given,

$f\left(x\right)=\frac{3x-1}{2},x\in R$

For one-one :

Let $x_1,x_2\in R$ such that $f\left(x_1\right)=f\left(x_2\right)$

$\frac{3x_1-1}{2}=\frac{3x_2-1}{2}$

$3x_1-1=3x_2-1$

$3x_1=3x_2$

$x_1=x_2$

Therefore, f is one-one.

For onto :

Let $y\in R$, then $f\left(x\right)=y$

$\frac{3x-1}{2}=y$

$3x-1=2y$

$x=\frac{2y+1}{3}\in R$

Thus, for each $y\in R$, there exists $x\in R$ such that

$f\left(\frac{2y+1}{3}\right)=y$

Hence, f is onto.

Therefore, f is bijective. Hence, $f^{-1}$ exists.

Now,

$x=\frac{2y+1}{3}$

$f^{-1}\left(y\right)=\frac{2y+1}{3}$

Therefore,

$f^{-1}\left(x\right)=\frac{2x+1}{3}$