Prove that


Let $f: R \rightarrow R: f(x)=\frac{1}{2}(3 x+1) .$ Show that $f$ is invertible and find $f^{-1}$.


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 $f: A \rightarrow B$ 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 $f(x)=\frac{(3 x+1)}{2}$ So $f\left(x_{1}\right)=f\left(x_{2}\right) \rightarrow \frac{\left(3 x_{1}+1\right)}{2}=\frac{\left(3 x_{2}+1\right)}{2} \rightarrow x_{1}=x_{2}$

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 R.

Let $y=f(x)=\frac{(3 x+1)}{2}$, So $x=\frac{2 y-1}{3}[$ Range of $f(x)=$ Domain of $y]$

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

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

So, $f(x)$ is onto function

As it is bijective function. So it is invertible

Invers of $f(x)$ is $f^{-1}(y)=\frac{2 y-1}{3}$



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