Chapter 4.3, Problem 10QR

To determine

To Find: The inverse of the given function.

Answer to Problem 10QR

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

Explanation of Solution

Given information: Given function $y=\arctan \frac{x}{3}$

Calculation:

First we have to prove that the given function $f(x)=\arctan \frac{x}{3}$ is one-one onto(bijective).

For one-one, we have

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

For onto, we have

  $\begin{array}{l}y=\arctan \frac{x}{3}\\ \tan y=\frac{x}{3}\\ x=3\tan y\end{array}$

Now, if $y\in \left(\frac{-\pi }{2},\frac{\pi }{2}\right)$ , then there exists $x=3\tan y\in R$ such that

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

Thus, $f$ is bijective and therefore invertible.

Now,

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