Chapter 4.6, Problem 17E

To determine

To sketch: the graph of two full period of the given function.

Explanation of Solution

Given information:

Given function is

  $y=-\frac{1}{2}\sec x$ .

Calculation:

The vertical asymptotes for $y=\sec x$ are at $x=-\left(\frac{\pi }{2}\right)$ and $x=\frac{\pi }{2}$ and its period is $2\pi$ .

So for the given equation, let us we find out the values of the two consecutive asymptotes occur at $x=-\left(\frac{\pi }{2}\right)$ and $x=\left(\frac{\pi }{2}\right)$ .

Next, calculate the values of $y$ foe a few values of $x$ between these two asymptotes.

For example,

Substitute $x=-\frac{\pi }{4}$ as shown:

  $\begin{array}{c}y=-\frac{1}{2}\sec \left(-\frac{\pi }{4}\right)\\ y=-\frac{1}{2}\left(\sqrt{2}\right)\\ y=-\frac{1}{\sqrt{2}}\end{array}$

Similarly, we calculate other values of $y$ and write them down in tabular form.

$x$	$-\left(\frac{\pi }{2}\right)$	$-\left(\frac{\pi }{4}\right)$	0	$\left(\frac{\pi }{4}\right)$	$\left(\frac{\pi }{2}\right)$
$y$	Not defined	$-\frac{1}{\sqrt{2}}$	$-\frac{1}{2}$	$-\frac{1}{\sqrt{2}}$	Not defined

Now using these values, we plot the graph including two full periods which means from $-2\pi$ to $2\pi$ .

  

The above graph represents the equation $y=-\frac{1}{2}\sec x$ .