Question

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

Answer 1

$f:R_{+}\to [4,\infty ]$ is given as $f\left(x\right)=x^2+4$

For one-one:

Let $f\left(x\right)=f\left(y\right)$

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

$x^2=y^2$

$x=y$ $[asx=y\in R_{+}]$

Therefore, f is a one-one function.

For onto:

For $yϵ[4,\infty ]$, let $y=x^2+4$

$x^2=y-4\ge 0$ [as$y\ge 4]$

$x=\sqrt{y-4}\ge 0$

Therefore, for any $y\in [4,\infty ]$ there exists $x=\sqrt{y-4}ϵR_{+}$, 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: $[4,\infty ]\to R_{+}$ by $g\left(y\right)=\sqrt{y-4}$

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

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

Therefore, $g\circ f=f\circ g=I_R$

Hence, f is invertible and the inverse of f is given by $f^{-1}\left(y\right)=g\left(y\right)=\sqrt{y-4}$