Chapter EP, Problem 7.2.18EP

To determine

To calculate: Whether the pair of functions $f\left(x\right)=3x-7\text{ and }g\left(x\right)=\frac{1}{3}x+7$ are inverse functions or not.

Answer to Problem 7.2.18EP

The pair of functions $f\left(x\right)=3x-7\text{ and }g\left(x\right)=\frac{1}{3}x+7$ are not inverse functions.

Explanation of Solution

Given information:

The functions $f\left(x\right)=3x-7\text{ and }g\left(x\right)=\frac{1}{3}x+7$ .

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

Calculation:

Consider the functions $f\left(x\right)=3x-7\text{ and }g\left(x\right)=\frac{1}{3}x+7$ .

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,

  $\begin{array}{c}\left[f\circ g\right]\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\frac{1}{3}x+7\right)\\ =3\left(\frac{1}{3}x+7\right)-7\\ =x+21-7\end{array}$

Simplify it further as,

  $\left[f\circ g\right]\left(x\right)=x+14$

And,

  $\begin{array}{c}\left[g\circ f\right]\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(3x-7\right)\\ =\frac{1}{3}\left(3x-7\right)+7\\ =x-\frac{7}{3}+7\end{array}$

Similarly,

  $\begin{array}{c}\left[g\circ f\right]\left(x\right)=x+\frac{21-7}{3}\\ =x+\frac{14}{3}\end{array}$

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

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

Thus, the pair of functions $f\left(x\right)=3x-7\text{ and }g\left(x\right)=\frac{1}{3}x+7$ are not inverse functions.