Solve this

Solution:

To Show: that $\mathrm{f}$ is invertible

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 $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ 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 $f(x)=3 x-4 .$ So $f\left(x_{1}\right)=f\left(x_{2}\right) \rightarrow 3 x_{1}-4=3 x_{2}-4 \rightarrow x_{1}=x_{2}$

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 $Q .$

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

So Domain of $y$ is $Q=$ Range of $f(x)$

Hence, Range of $f(x)=$ co-domain of $f(x)=Q$

So, $f(x)$ is onto function

As it is bijective function. So it is invertible

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