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

$f:R_1\to (4,\infty )$

$f\left(x\right)=x^2+4$

$f\left(x_1\right)=f\left(x_2\right)$

${x_1}^2+4={x_2}^2+4$

$x_1=x_2$ $f\left(x\right)$ is on-one.

$f\left(x\right)=x^2+4$

$y=x^2+4$

$x^2=y-4$

$x=\sqrt{y-4}$

$f\left(x\right)=x^2+4$

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

$f\left(x\right)=y-4+4$

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

$f\left(x\right)$ is onto function.

So,$f\left(x\right)$ is bijective, Hence it is invertible

$f\left(x\right)=x^2+4$

$y=x^2+4$

$y=x^2-4$

$y=\sqrt{x^2-4}$

$f^{-1}\left(x\right)=\sqrt{x-4}$