Help me with precal pleaseee

Answer:  Choice C

[tex]\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right][/tex]

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Explanation:

Let's look at the function y = sec(x) first, which is the secant function.

Recall that secant is 1 over cosine, so sec(x) = 1/cos(x)

We can't divide by zero, so cos(x) = 0 can't be allowed. If x = pi/2, then cos(pi/2) = 0 will happen. So we must exclude pi/2 from the domain of sec(x).

If we look at the interval from 0 to pi, then the domain of sec(x) is [tex]0 \le x < \frac{\pi}{2} \ \ \text{ and } \ \ \frac{\pi}{2} < x \le \pi[/tex]

we can condense that into the interval notation [tex]\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right][/tex]

Note the use of curved parenthesis to exclude the endpoint; while the square bracket includes the endpoint.

So effectively we just poked at hole at x = pi/2 to kick that out of the domain. I'm only focusing on the interval from 0 to pi so that secant is one to one on this interval. That way we can apply the inverse. When we apply the inverse, the domain and range swap places. So the range of arcsecant, or [tex]\sec^{-1}(x)[/tex] is going to also be [tex]\left[0 , \frac{\pi}{2}\right) \ \ \text{ and } \ \ \left(\frac{\pi}{2}, \pi\right][/tex]

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