# Question #f8bc2

Dec 18, 2015

$y = {\sec}^{- 1} \sqrt{x}$ is undefined for all real $x | x < 1$.

#### Explanation:

There's a way to rewrite $y$ in a way that's a little easier to analyse.

First, as a small aside, let's say we just have the secant function

$z \left(\theta\right) = \sec \theta$.

The inverse, ${\sec}^{-} 1 \theta$ we can obtain pretty easily. We'll start here, with

$\theta = \sec \left({z}^{- 1} \left(\theta\right)\right)$.

Now let's actually set $\sec \left({z}^{-} 1 \left(\theta\right)\right)$ equal to $\frac{1}{\cos} \left({z}^{-} 1 \left(\theta\right)\right)$. Then

$\theta = \frac{1}{\cos} \left({z}^{-} 1 \left(\theta\right)\right)$.

Just a little algebra and we've solved for ${z}^{-} 1$:

${z}^{-} 1 \left(\theta\right) = {\cos}^{-} 1 \left(\frac{1}{\theta}\right)$

So, really what we've just demonstrated is that the inverse of $z = \sec \theta$, also known as arc secant of $\theta$, is equal to ${\cos}^{-} 1 \left(\frac{1}{\theta}\right)$.

How does this help us? Well, for one thing, we can rewrite our little function in terms of $\cos$ instead of $\sec$.

Take a look:

$y = {\sec}^{- 1} \sqrt{x} = {\cos}^{-} 1 \left(\frac{1}{\sqrt{x}}\right)$

We already know that the inverse cosine function ${\cos}^{-} 1 x$ is only defined for $- 1 \le x \le 1$. So, what this should tell us is that $y$ will only be defined for

$- 1 \le \frac{1}{\sqrt{x}} \le 1$

Multiply through by $\sqrt{x}$ and we obtain

$- \sqrt{x} \le 1 \le \sqrt{x}$.

Let's split this into two different inequalities so it's a little easier to figure out:

$1 \ge - \sqrt{x}$
$1 \le \sqrt{x}$

Now, multiply the first one through by $- 1$.

$- 1 \le \sqrt{x}$
$1 \le \sqrt{x}$

And square both sides of each inequality:

$1 \le x$
$1 \le x$

We've reduced our condition to one inequality now, and the discontinuity should be pretty obvious at this point. The function is only defined for $x \ge 1$.