Prove that

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 $f: A \rightarrow B$ 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 $x_{1}, x_{2} \in R$ and $f(x)=\left(9 x^{2}+6 x-5\right) .$ So $f\left(x_{1}\right)=f\left(x_{2}\right) \rightarrow\left(9 x_{1}^{2}+6 x_{1}-5\right)=\left(9 x_{2}^{2}+6 x_{2}-5\right)$ on solving we get $\rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$

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

Let $y=f(x)=\left(9 x^{2}+6 x-5\right)$, So $x=\frac{-1+\sqrt{y+6}}{3}[$ Range of $f(x)=$ Domain of $y]$

So Domain of $y=$ Range of $f(x)=[-5, \infty]$

Hence, Range of $f(x)=$ co-domain of $f(x)=[-5, \infty]$

So, $f(x)$ is onto function

As it is bijective function. So it is invertible

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