Q 42. - Precalculus Enhanced with Graphing Utilities

Question

Verify that the functions $f\left(x\right)=\frac{x-5}{2x+3}$ and $g\left(x\right)=\frac{3x+5}{1-2x}$ 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 .Exclude $x=-\frac{3}{2}$ from domain of f and $x=\frac{1}{2}$ from domain of g.

Step 1. Given information.

Given functions $f\left(x\right)=\frac{x-5}{2x+3}$ and $g\left(x\right)=\frac{3x+5}{1-2x}$.

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.

$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& f\left(\frac{3x+5}{1-2x}\right)\\ & =& \frac{\left(\frac{3x+5}{1-2x}\right)-5}{2\left(\frac{3x+5}{1-2x}\right)+3}\\ & =& \frac{3x+5-5+10x}{6x+10+3-6x}\\ & =& \frac{13x}{13}\\ & =& x\end{array}$

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

$\begin{array}{rcl}g\left(f\left(x\right)\right)& =& g\left(\frac{x-5}{2x+3}\right)\\ & =& \frac{3\left(\frac{x-5}{2x+3}\right)+5}{1-2\left(\frac{x-5}{2x+3}\right)}\\ & =& \frac{3x-15+10x+15}{2x+3-2x+10}\\ & =& \frac{13x}{13}\\ & =& x\end{array}$

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 $x\ne -\frac{3}{2}$ and that of g is $x\ne \frac{1}{2}$.

It follows that f exists for all real numbers except for $-\frac{3}{2}$ and g exists for all real numbers except for $\frac{1}{2}$.

Therefore,  $x=-\frac{3}{2}$ and $x=\frac{1}{2}$need to be excluded from domain of f and g respectively.