Question

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

Answer 1

Here, function $f:R_{+}\to [4,\infty )$ is given $f\left(x\right)=x^2+4$
Let $x,y\in R_{+}$, such that $f\left(x\right)=f\left(y\right)\Rightarrow x^2+4=y^2+4$ $\Rightarrow x^2=y^2\Rightarrow x=y[asx=y\in R_{+}]$
Therefore, f is a one-one function.
For $y\in [4,\infty )$, let $y=x^2+4$
$\Rightarrow x^2=y-4\ge 0[asy\ge 4]$

Therefore, for any $y\in R_{+}$, There exists $x=\sqrt{y-4}\in R_{+}$ such that
$f\left(x\right)=f\left(\sqrt{y-4}\right)=(\sqrt{y-4}{)}^2+4=y-4+4=y$
Therefore, f is onto, Thus, f is one-one and onto and therefore, $f^{-1}$ exists.
Let us define $g:[4,\infty )\to R_{+}byg\left(y\right)=\sqrt{y-4}$
Now, $gof\left(x\right)=g\left(f\right(x\left)\right)$=$g(x^2+4)$=$\sqrt{(x^2+4)-4}$=$\sqrt{x^2}=x$
and $fog\left(y\right)=f\left(g\right(y\left)\right)=f\left(\sqrt{y-4}\right)=(\sqrt{y-4}{)}^2+4=(y-4)+4=y$
Therefore, $gof=I_{R_{+}}andfog=I_{4,\infty }$
$f^{-1}\left(y\right)=g\left(y\right)=\sqrt{y-4}$