Question

Question #f8bc2

Answer 1

`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(theta) = sec theta`.

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

`theta = sec (z^(-1)(theta))`.

Now let's actually set `sec (z^-1(theta))` equal to `1/cos(z^-1(theta))`. Then

`theta = 1/cos(z^-1(theta))`.

Just a little algebra and we've solved for `z^-1`:

`z^-1(theta) = cos^-1 (1/theta)`

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 (1/theta)`.

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 (1/sqrt(x))`

We already know that the inverse cosine function `cos ^-1 x` is only defined for `-1 <= x <= 1`. So, what this should tell us is that `y` will only be defined for

`-1 <= 1/sqrt(x) <= 1`

Multiply through by `sqrt(x)` and we obtain

`-sqrt(x) <= 1 <= sqrt(x)`.

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

`1 >= -sqrt(x)`
`1 <= sqrt(x)`

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

`-1 <= sqrt(x)`
`1 <= sqrt(x)`

And square both sides of each inequality:

`1 <= x`
`1 <= 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 >= 1`.