Question

Prove that the function $f:N\to N,$ defined by $f\left(x\right)=x^2+x+1$ is one-one but not onto. Find inverse of $f:N\to S$, where $S$ is range of $f.$

Answer 1

$\text{Given}f\left(x\right)=x^2+x+1$
To prove one-one, we will show if $f\left(x_1\right)=f\left(x_2\right),$ then $x_1=x_2$
$\text{Let}f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow x_1^2+x_1+1=x_2^2+x_2+1$
$\Rightarrow x_1^2+x_1-x_2^2-x_2=0$
$\Rightarrow (x_1-x_2)(x_1+x_2+1)=0$
$\text{Since}{x}_1+x_2+1\ne 0\forall x_1,x_2\in N,$
we have $x_1-x_2=0$
$\Rightarrow x_1=x_2$
$\text{Thus,}f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$
​​​​​​Hence, $f$ is one-one.

$f\left(x\right)=1$ has no pre-image in $R.$
So, $f$ is not onto.

$f:N\to S$ is one- one and onto.
Hence, $f:N\to S$ is invertible.
To find the inverse of $f$
$\text{Let}f\left(x\right)=y$
$\Rightarrow x=f^{-1}\left(y\right)\cdots \left(1\right)$
$\text{and}y=x^2+x+1$
$\Rightarrow y={(x+\frac{1}{2})}^2+\frac{3}{4}$
$\Rightarrow \sqrt{y-\frac{3}{4}}=x+\frac{1}{2}$
(Taking positive root only because $x\in N$ and hence, RHS is always positive)
$\Rightarrow \sqrt{y-\frac{3}{4}}-\frac{1}{2}=x$
$\Rightarrow \sqrt{y-\frac{3}{4}}-\frac{1}{2}=f^{-1}\left(y\right)\left[\text{From}\right(1\left)\right]$
On replacing $y$ with $x$, we get
$f^{-1}\left(x\right)=\sqrt{x-\frac{3}{4}}-\frac{1}{2}$