Chapter 4.6, Problem 13E

To determine

To state: the period of the function and match with its graph.

Answer to Problem 13E

Graph of $y=\frac{1}{2}\sec \frac{\pi x}{2}$ is graph (f).

  $y=\frac{1}{2}\sec \frac{\pi x}{2}$ has period 4.

Explanation of Solution

Given information:

Given function is

  $y=\frac{1}{2}\sec \frac{\pi x}{2}$ .

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 $2p$ .

So for the given equation, let us we find out the values of the two consecutive asymptotes.

On comparing we get,

  $\begin{array}{c}\frac{\pi x}{2}=-\frac{\pi }{2}\\ x=\left(\frac{2}{\pi }\right)\left(-\frac{\pi }{2}\right)\\ x=-1\end{array}$

Also,

  $\begin{array}{c}\frac{\pi x}{2}=\frac{\pi }{2}\\ x=\left(\frac{\pi }{2}\right)\left(\frac{2}{\pi }\right)\\ x=1\end{array}$

Therefore, two consecutive asymptotes occur at $x=-1$ and $x=1$ .

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

For example,

Substituting $x=-\frac{1}{2}$

We get,

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

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

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

Now, we will plot the graph for the following values form $-4$ to 4

  

The above graph represents the equation $y=\frac{1}{2}\sec \left(\frac{\pi x}{2}\right)$ .

Now, as the above graph resembles the graph given in option $\left(f\right)$ .

Also, from the above graph we can see that the function $y=\frac{1}{2}\sec \left(\frac{\pi x}{2}\right)$ complete its one cycle from $-1$ to 3. Thus, the period of the function is 4.

Therefore, the correct option is $\left(f\right)$ and the period of the function is 4.