Chapter 4.7, Problem 76E

To determine

Identify any asymptotes of the graphs.

Answer to Problem 76E

  $\boxed{x=0}$

Explanation of Solution

Given information:

Use a graphing utility to graph f and g in the same viewing window to verify that the two functions are equal. Explain why they are equal. Identify any asymptotes of the graphs.

  $f\left(x\right)=\tan \left(\arccos \frac{x}{2}\right),g\left(x\right)=\frac{\sqrt{4-x^2}}{x}$

Calculation:

Consider the functions

  $f\left(x\right)=\tan \left(\arccos \frac{x}{2}\right)$ ......(1)

  $g\left(x\right)=\frac{\sqrt{4-x^2}}{x}$ .....(2)

According to the following Inverse Trigonometric Functions

  $y=\arccos x$ , if and only if $\cos y=x$ defined in $\left[0,\pi \right]$

It is know that

  $\sin \theta =\frac{\text{opp}}{\text{hyp}}$

  $\cos \theta =\frac{adj}{hyp}$

  $\begin{array}{l}\tan \theta =\frac{\text{opp}}{adj}\\ \end{array}$

Use $Ti-83$ Calculator to plot the graph of the above two functions in the same viewing window.

Start $Ti-83$ Calculator and press $\boxed{Y=}$ and input the following functions.

  $\begin{array}{l}{Y}_{1=}\tan \left({\cos }^{-1}\left(X/2\right)\right)\\ Y_2=\sqrt{}\left(4-X^2\right)/X\end{array}$

This display is as shown below

  

Press the $[GRAPH]$ button to get the graph as shown below

  

As the graph of $Y_1$ and $Y_2$ overlap each other the function $f$ and $g$ are equal.

Explanation of equality of the function $f$ and $g$

Consider the figure made with reference to (1)

In triangle $ABC$ , from Pythagoras Theorem,

  $A{C}^2=A{B}^2+B{C}^2$

That is

  $AB=\sqrt{A{C}^2-B{C}^2}$

  $AB=\sqrt{2^2-x^2}$

  $AB=\sqrt{4-x^2}$

In the above figure,

  $\tan \theta =\frac{opp}{adj}$

  $\tan \theta =\frac{\sqrt{4-x^2}}{x}$

  $\theta =\arctan \frac{\sqrt{4-x^2}}{x}$

Put $\theta$ as the argument in(1), the expression becomes

  $\tan \left(\arctan \frac{\sqrt{4-x^2}}{x}\right)$

Using properties of inverse trigonometric functions

  $\tan \left(\arctan \frac{\sqrt{4-x^2}}{x}\right)=\frac{\sqrt{4-x^2}}{x}$

And this is equal to (2) Hence verified that function $f$ and $g$ are equal

In (2) since the denominator of the function $f$ is zero at $x=0$

That is

  $\lim g(x)=\lim \frac{\sqrt{4-x^2}}{x}$

  $=\frac{\sqrt{4-x^2}}{0}$

  $=\frac{\sqrt{4-x^2}}{0}$

  $=\infty$

Hence, the line $\boxed{x=0}$ is a vertical asymptote to the graph of a function $f$ and $g$ .