Chapter EP, Problem 7.2.17EP

To determine

To calculate: Whether the pair of functions $f\left(x\right)=6-3x\text{ and }g\left(x\right)=2-\frac{1}{3}x$ are inverse functions or not.

Answer to Problem 7.2.17EP

The pair of functions $f\left(x\right)=6-3x\text{ and }g\left(x\right)=2-\frac{1}{3}x$ are inverse functions.

Explanation of Solution

Given information:

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

Formula used:

The composition of two functions f and gis $\left[f\circ g\right]\left(x\right)=f\left(g\left(x\right)\right)$ .

The two functions f and g are said to be inverse functions if and only if $\left[f\circ g\right]\left(x\right)=\left[g\circ f\right]\left(x\right)=x$ .

Calculation:

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

Recall that the composition of two functions f and gis $\left[f\circ g\right]\left(x\right)=f\left(g\left(x\right)\right)$ .

Therefore, $\left[f\circ g\right]\left(x\right)\text{ and }\left[g\circ f\right]\left(x\right)$ are evaluated as,

  $\begin{array}{c}\left[f\circ g\right]\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(2-\frac{1}{3}x\right)\\ =6-3\left(2-\frac{1}{3}x\right)\\ =6-6+x\end{array}$

Simplify it further as,

  $\left[f\circ g\right]\left(x\right)=x$

And,

  $\begin{array}{c}\left[g\circ f\right]\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(6-3x\right)\\ =2-\frac{1}{3}\left(6-3x\right)\\ =2-2+x\end{array}$

Similarly,

  $\left[g\circ f\right]\left(x\right)=x$

Recall that the two functions f and gare said to be inverse functions if and only if $\left[f\circ g\right]\left(x\right)=\left[g\circ f\right]\left(x\right)=x$ .

Since, $\left[f\circ g\right]\left(x\right)=\left[g\circ f\right]\left(x\right)=x$ .

Thus, the pair of functions $f\left(x\right)=6-3x\text{ and }g\left(x\right)=2-\frac{1}{3}x$ are inverse functions.