Show that the function

Question:

Show that the function $f$ on $A=R-\left\{\frac{2}{3}\right\}$, defined as $f(x)=\frac{4 x+3}{6 x-4}$ is one-one and onto. Hence, find $f^{-1}$.

Solution:

To Show: that $\mathrm{f}$ is one-one and onto

To Find: Inverse of $f$

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function $f: A \rightarrow B$ is said to be a one-one function or injective mapping if different

elements of A have different images in B. Thus for $x_{1}, x_{2} \in A \& f\left(x_{1}\right), f\left(x_{2}\right) \in B, f\left(x_{1}\right)=f\left(x_{2}\right) \leftrightarrow x_{1}=x_{2}$ or $x_{1} \neq$ $x_{2} \leftrightarrow f\left(x_{1}\right) \neq f\left(x_{2}\right)$

onto function: If range $=$ co-domain then $f(x)$ is onto functions.

So, We need to prove that the given function is one-one and onto.

Let $x_{1}, x_{2} \in Q$ and $f(x)=\frac{(4 x+3)}{(6 x-4)} .$ So $f\left(x_{1}\right)=f\left(x_{2}\right) \rightarrow \frac{\left(4 x_{1}+3\right)}{\left(6 x_{1}-4\right)}=\frac{\left(4 x_{2}+3\right)}{\left(6 x_{2}-4\right)} \rightarrow$ on solving we get $x_{1}=x_{2}$

So $f\left(x_{1}\right)=f\left(x_{2}\right) \leftrightarrow x_{1}=x_{2}, f(x)$ is one-one

Given co-domain of $f(x)$ is $R$ except $3 x-2=0$.

Let $y=f(x)=\frac{(4 x+3)}{(6 x-4)}$ So $x=\frac{4 y+3}{6 y-4}[$ Range of $f(x)=$ Domain of $y]$

So Domain of $y$ is $R$ (except $3 x-2=0$ ) $=$ Range of $f(x)$

Hence, Range of $f(x)=$ co-domain of $f(x)=R$ except $3 x-2=0$

So, $f(x)$ is onto function

As it is bijective function. So it is invertible

Invers of $f(x)$ is $f^{-1}(y)=\frac{4 y+3}{6 y-4}$