Chapter 4.3, Problem 6QR

To determine

To Find: The inverse of the given function.

Answer to Problem 6QR

The inverse is $f^{-1}(x)=\frac{x+8}{3}$

Explanation of Solution

Given information: Given function $y=3x-8$

Calculation:

First we have to prove that the given function $f(x)=3x-8$ is one-one onto(bijective).

For one-one, we have

  $\begin{array}{l}f(x_1)=f(x_2)\\ 3x_1-8=3x_2-8\\ 3x_1=3x_2\\ x_1=x_2\\ \therefore f\text{is}\text{one-one}\text{.}\end{array}$

For onto, we have

  $\begin{array}{l}y=3x-8\\ 3x=y+8\\ x=\frac{y+8}{3}.\end{array}$

Now, if $y\in R$ , then there exists $x=\frac{y+8}{3}\in R$ such that

  $\begin{array}{l}f(x)=f\left(\frac{y+8}{3}\right)=3\left(\frac{y+8}{3}\right)-8=y\\ \therefore f\text{is}\text{onto}\text{.}\end{array}$

Thus, $f$ is bijective and therefore invertible.

Now,

  $\begin{array}{l}f\circ f^{-1}(x)=x\\ f(f^{-1}(x))=x\\ 3f^{-1}(x)-8=x\\ \Rightarrow f^{-1}(x)=\frac{x+8}{3}\end{array}$