Q 38. - Precalculus Enhanced with Graphing Utilities

Question

Verify that the functions $f\left(x\right)={\left(x-2\right)}^2$, $x\ge 2$ and $g\left(x\right)=\sqrt{x}+2$ are inverses of each other by showing that $f\left(g\left(x\right)\right)=x$ and $g\left(f\left(x\right)\right)=x$. Give any values of x that need to be excluded from the domain of f and the domain of g.

Step-by-Step Solution

Answer

Functions f and g are inverses of one another . The value of x need to be excluded from domain of f and g is $-1$.

Step 1. Given information.

Given functions $f\left(x\right)={\left(x-2\right)}^2,x\ge 2$ and $g\left(x\right)=\sqrt{x}+2$.

Step 2. Verify that the functions f and g and are inverses of each other.

Note that functions f and g are inverses of each other if $f\left(g\left(x\right)\right)=x$ and $g\left(f\left(x\right)\right)=x$.

Prove $f\left(g\left(x\right)\right)=x$ as follows.

$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& f\left(\sqrt{x}+2\right)\\ & =& {\left(\sqrt{x}+2-2\right)}^2\\ & =& {\left(\sqrt{x}\right)}^2\\ & =& x\end{array}$

Prove $g\left(f\left(x\right)\right)=x$ as follows.

 $\begin{array}{rcl}g\left(f\left(x\right)\right)& =& g\left({\left(x-2\right)}^2\right)\\ & =& \sqrt{{\left(x-2\right)}^2}+2\\ & =& x-2+2\\ & =& x\end{array}$

Therefore, f and g are inverses of each other.

Step 3. Find values of x that need to be excluded from domain of f and g .

Note that domain of f is $x\ge 2$and that of g are all real number.

It follows that both function does not exists for $x=-1$.

Therefore,  $x=-1$ need to be excluded from domain of both the functions.