The function $f$is one-to-one. Find the inverse of each function and check your answer. State the

domain and the range of $f$and $f^{-1}$.

$f\left(x\right)=\sqrt{x-2}$

At first, in $y=f\left(x\right)$, interchange the variables role="math" localid="1646442857368" $x$and role="math" localid="1646442851447" $y$, to obtain $x=f\left(y\right)$.

Then solve the equation for role="math" localid="1646442871400" $y$in terms of role="math" localid="1646442875781" $x$.

After that, check the result by showing that,

$f^{-1}\left(f\left(x\right)\right)=x$androle="math" localid="1646442886071" $f\left(f^{-1}\left(x\right)\right)=x$

Given that, $f\left(x\right)=\sqrt{x-2}$

Replacing $f\left(x\right)$withrole="math" localid="1646443190865" $y$and interchange the variables $x$and $y$in

$y=\sqrt{x-2}$

to obtain

$x=\sqrt{y-2}$

$x=\sqrt{y-2}$

Squaring both sides,

$x^2=y-2$

Add $2$with both sides,

$x^2+2=y$

$y=x^2+2$

The inverse is

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

$f^{-1}\left(f\left(x\right)\right)=f^{-1}\left(\sqrt{x-2}\right)$

$={\left(\sqrt{x-2}\right)}^2+2$

$=x-2+2$

$=x$

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

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

$=\sqrt{x^2}$

$=x$

Thus, the result is checked.

$f^{-1}\left(f\left(x\right)\right)=x$, $x$is in domain of $f$and $Domainoff=Rangeof{f}^{-1}$

$f\left(f^{-1}\left(x\right)\right)=x$, $x$is in domain of localid="1646449702556" $f^{-1}$and $Domainof{f}^{-1}=Rangeoff$