Question

If the function $f\left(x\right)=\sqrt{2x-3}$ is invertible then find its inverse. Hence prove that $\left(fo{f}^{-1}\right)\left(x\right)=x$.

Answer 1

Let $y=\sqrt{2x-3}$

As $f\left(x\right)$ is invertible so Domain of $f\left(x\right)$ will be $2x-3\ge 0\to x\ge \frac{3}{2}$

We need to find inverse of $y$.

Squaring on both sides, we get,

$y^2=2x-3\to x=\frac{y^2+3}{2}$

Hence $f^{-1}\left(x\right)=\frac{x^2+3}{2}$

Now, $\left(fo{f}^{-1}\right)\left(x\right)=f\left(f^{-1}\right(x\left)\right)=\sqrt{2\left(f^{-1}\right(x)-3})=\sqrt{2\left(\frac{x^2+3}{2}\right)-3}$

$=\sqrt{x^2+3-3}=\sqrt{x^2}=|x|$

Now as obtained from domain $x\ge \frac{3}{2}$

$\therefore |x|=x$

So, $\left(fo{f}^{-1}\right)\left(x\right)=f\left(f^{-1}\right(x\left)\right)=x$

Hence proved.