Question

Consider a function $f:R^{+}\to (-1,\infty ]$ defined as $f\left(x\right)=5x^2+2x-1$. Then find whether $f$ is invertible or not and if it is then find $f^{-1}\left(x\right)$

Answer 1

Given $f:R^{+}\to (-1,\infty ]$
$f\left(x\right)=5x^2+2x-1$

For $f$ is to invertible, it should be both one-one and onto function.

For one-one,
$f\left(x_1\right)=f\left(x_2\right)$
$\Rightarrow 5x_1^2+2x_1-1=5x_2^2+2x_2-1\Rightarrow 5(x_1^2-x_2^2)+2(x_1-x_2)=0$
$\Rightarrow (x_1-x_2)(5x_1+5x_2+2)=0$
$\Rightarrow x_1=x_2$
$\because 5x_1+5x_2+2\ne 0$ as $x_1,x_2\in R^{+}$
So, $f$ is one-one.

For onto,
$y=5x^2+2x-1$
$\Rightarrow \frac{y+1}{5}=x^2+\frac{2}{5}x\Rightarrow \frac{y+1}{5}={(x+\frac{1}{5})}^2-\frac{1}{25}\Rightarrow \frac{5y+6}{25}={(x+\frac{1}{5})}^2\Rightarrow \sqrt{\frac{5y+6}{25}}=(x+\frac{1}{5})\Rightarrow \sqrt{\frac{5y+6}{25}}-\frac{1}{5}=x\Rightarrow x=\frac{\sqrt{5y+6}-1}{5}$
$\therefore \forall y\in (-1,\infty ],$ the domain of $f\left(x\right)$ is $x\in R^{+}$.
So, $f\left(x\right)$ is onto.

Since, $f$ is both one-one and onto. So, it is invertible.

$f^{-1}\left(y\right)=\frac{\sqrt{5y+6}-1}{5}$
$\therefore f^{-1}\left(x\right)=\frac{1}{5}(\sqrt{5x+6}-1)$