Q 36. - Precalculus Enhanced with Graphing Utilities

Question

Verify that the functions $f\left(x\right)=2x+6$ and $g\left(x\right)=\frac{1}{2}x-3$ 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 . No value of x need to be excluded from domain of f and g.

Step 1. Given information.

Given functions $f\left(x\right)=2x+6$ and $g\left(x\right)=\frac{1}{2}x-3$.

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.

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

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

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

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 both f and g are all real number.

It follows that both function exists for all real numbers.

Therefore, no value of x need to be excluded from domain of both the functions.