# Q. 8 Consider $f : R_+\rightarrow [4,\infty)$ given by $f (x) = x^2 + 4$. Show that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1} = \sqrt{y-4}$ , where $R_+$ is the set of all non-negative real numbers.

S seema garhwal

$f : R_+\rightarrow [4,\infty)$

$f (x) = x^2 + 4$

One- one:

Let       $f(x)=f(y)$    $for \, \, x,y\in R$

$x^{2}+4=y^{2}+4$

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

$\Rightarrow$              $x=y$

$\therefore$   f is one-one.

Onto:

Let for  $y \in [4,\infty)$  , $y=x^{2}+4$

$\Rightarrow$         $x^{2}-y+4\geq 0$

$\Rightarrow$             $x^{2}\geq y-4$

$\Rightarrow$                  $x\geq (y-4)^{\frac{1}{2}}$

$\therefore$ for $y \in [4,\infty)$ there is $x= {\sqrt{(y-4)}}$  such that

$f(x)= f(\sqrt{y-4})= \left ( \sqrt{y-4} \right )^{2}+4$

$= y-4 +4 = y$

$\therefore$ f is onto.

Since, f is one-one and onto so it is invertible.

Let  $g : [4,\infty)\rightarrow R_+$    by  $g(y)= {\sqrt{(y-4)}}$

Now,  $(gof)(x)=g(f(x))=g(x^{2}+4)=\sqrt{x^{2}-4+4}=\sqrt{x^{2}}=x$

$(fog)(x)=f(g(x))=f(\sqrt{y-4})=(\sqrt{y-4})^{2}+4=y-4+4=y$

$\therefore gof=fog=I_R$

Hence,  $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1} = \sqrt{y-4}$

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