Show that the function f : N → N defined by

Given: The function $f: N \rightarrow N$ defined by $f(x)=x^{2}+x+1$

To show f is one-one:

Let $f\left(x_{1}\right)=f\left(x_{2}\right)$

$\Rightarrow x_{1}^{2}+x_{1}+1=x_{2}^{2}+x_{2}+1$

$\Rightarrow x_{1}^{2}+x_{1}=x_{2}^{2}+x_{2}$

$\Rightarrow x_{1}^{2}+x_{1}-x_{2}^{2}-x_{2}=0$

$\Rightarrow x_{1}^{2}-x_{2}^{2}+x_{1}-x_{2}=0$

$\Rightarrow\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)+\left(x_{1}-x_{2}\right)=0$

$\Rightarrow\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}+1\right)=0$

$\Rightarrow x_{1}-x_{2}=0$ or $x_{1}+x_{2}+1=0$

$\Rightarrow x_{1}=x_{2}$ or $x_{1}=-\left(x_{2}+1\right)$

$\Rightarrow x_{1}=x_{2} \quad\left(\because x_{1}, x_{2} \in N\right)$

Hence, $f$ is one - one.

To show $f$ is not onto:

Since $f(x)=x^{2}+x+1$

$\therefore f(1)=3$

$f(2)=7$

$f(3)=13$

and so on

Thus, Range of $f=\{3,7,13, \ldots\} \neq N$

Hence, $f$ is not onto.

Now, Let $f: N \rightarrow$ Range of $f$

$y=x^{2}+x+1$

$\Rightarrow x^{2}+x+1-y=0$

$\Rightarrow x^{2}+x+(1-y)=0$

$\Rightarrow x=\frac{-1 \pm \sqrt{1^{2}-4(1)(1-y)}}{2(1)}$

$\Rightarrow x=\frac{-1 \pm \sqrt{1-4+4 y}}{2}$

$\Rightarrow x=\frac{-1 \pm \sqrt{4 y-3}}{2}$

$\Rightarrow x=\frac{-1+\sqrt{4 y-3}}{2}$ or $x=\frac{-1-\sqrt{4 y-3}}{2}$

$\Rightarrow x=\frac{-1+\sqrt{4 y-3}}{2} \quad(\because x \in N)$

Hence, $f^{-1}(x)=\frac{-1+\sqrt{4 x-3}}{2} .$