Chapter 4.7, Problem 112E

To determine

The inverse secant function by restricting the domain of the secant function to the intervals $\left[0,\pi /2\right),\left(\pi /2,\pi \right]$ .

Answer to Problem 112E

The domain of $y={\sec }^{-1}x\text{is }\left[-\infty ,-1\right)\cup \left[1,\infty \right)$ and the range $y={\sec }^{-1}x\text{is }\left[0,\pi /2\right)\cup \left(\pi /2\pi \right]$

Explanation of Solution

Given information:

The interval value is given as $\left[0,\pi /2\right),\left(\pi /2,\pi \right]$

Formula used:

The graph is plotted against x- axis and y-axis.

Calculation:

If one restricts to the domain to the interval, the following properties hold in accordance to the secant function

1. On the interval $\left[0,\pi /2\right)\cup \left(\pi /2,\pi \right]$ the secant function is increasing

2. The secant function is not defined for the value $x=\frac{\pi }{2}$

3. On the interval $\left[0,\pi /2\right)\cup \left(\pi /2,\pi \right],y=\sec x$ takes on full range values,

4. On the interval $\left[0,\pi /2\right)\cup \left(\pi /2,\pi \right],y=\sec x$ is one-to-one

So, on the restricted domain $\left[0,\pi /2\right)\cup \left(\pi /2,\pi \right],y=\sec x$ has a unique inverse function called the inverse secant function?

It is denoted by

  $y=\mathrm{arcsec}x\text{or}y={\sec }^{-1}x$

If and only if sec y =x

Where $\left(-\infty ,-1\right]\cup \left[1,\infty \right)\text{and }\left[0,\pi /2\right)\cup \left(\pi /2,\pi \right]$

The domain of $y={\sec }^{-1}x\text{is }\left[-\infty ,-1\right)\cup \left[1,\infty \right)$

And the range of $y={\sec }^{-1}x\text{is }\left[0,\pi /2\right)\cup \left(\pi /2\pi \right]$

Graph of the secant function is shown below,

  

Conclusion:

The domain of $y={\sec }^{-1}x\text{is }\left[-\infty ,-1\right)\cup \left[1,\infty \right)$ and the range $y={\sec }^{-1}x\text{is }\left[0,\pi /2\right)\cup \left(\pi /2\pi \right]$