Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine which x-values are vertical asymptotes of the function [tex]\( y = \sec(x) \)[/tex], we need to understand where the secant function becomes undefined. The secant function is defined as:
[tex]\[ y = \sec(x) = \frac{1}{\cos(x)} \][/tex]
The secant function becomes undefined wherever [tex]\(\cos(x) = 0\)[/tex], because division by zero is undefined. Therefore, we need to find the x-values where [tex]\(\cos(x) = 0\)[/tex].
The cosine function, [tex]\(\cos(x)\)[/tex], equals zero at:
[tex]\[ x = \frac{(2n+1)\pi}{2} \][/tex]
where [tex]\( n \)[/tex] is any integer. This is because the cosine function has zeros at every odd multiple of [tex]\(\frac{\pi}{2}\)[/tex].
Let’s check each of the given options to see if they correspond to one of these points.
1. [tex]\( x = -2\pi \)[/tex]
- For [tex]\( x = -2\pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-2\pi = \frac{(2 \cdot -2 + 1)\pi}{2} = \frac{-3\pi}{2}\)[/tex], which is not correct because [tex]\(-3\pi/2\)[/tex] is not equivalent to [tex]\(-2\pi\)[/tex].
- So, [tex]\( x = -2\pi \)[/tex] is not an asymptote.
2. [tex]\( x = -\frac{\pi}{6} \)[/tex]
- For [tex]\( x = -\frac{\pi}{6} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-\frac{\pi}{6}\)[/tex] cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = -\frac{\pi}{6} \)[/tex] is not an asymptote.
3. [tex]\( x = \pi \)[/tex]
- For [tex]\( x = \pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\pi = \frac{2\pi}{2}\)[/tex], which cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = \pi \)[/tex] is not an asymptote.
4. [tex]\( x = \frac{3\pi}{2} \)[/tex]
- For [tex]\( x = \frac{3\pi}{2} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\frac{3\pi}{2} = \frac{(2 \cdot 1 + 1)\pi}{2}\)[/tex], which matches the form [tex]\(\frac{(2n+1)\pi}{2}\)[/tex] for [tex]\( n = 1 \)[/tex].
- Therefore, [tex]\( x = \frac{3\pi}{2} \)[/tex] is an asymptote.
Therefore, the correct answer is:
[tex]\[ x = \frac{3\pi}{2} \][/tex]
The option that corresponds to this is:
[tex]\[ 4 \][/tex]
[tex]\[ y = \sec(x) = \frac{1}{\cos(x)} \][/tex]
The secant function becomes undefined wherever [tex]\(\cos(x) = 0\)[/tex], because division by zero is undefined. Therefore, we need to find the x-values where [tex]\(\cos(x) = 0\)[/tex].
The cosine function, [tex]\(\cos(x)\)[/tex], equals zero at:
[tex]\[ x = \frac{(2n+1)\pi}{2} \][/tex]
where [tex]\( n \)[/tex] is any integer. This is because the cosine function has zeros at every odd multiple of [tex]\(\frac{\pi}{2}\)[/tex].
Let’s check each of the given options to see if they correspond to one of these points.
1. [tex]\( x = -2\pi \)[/tex]
- For [tex]\( x = -2\pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-2\pi = \frac{(2 \cdot -2 + 1)\pi}{2} = \frac{-3\pi}{2}\)[/tex], which is not correct because [tex]\(-3\pi/2\)[/tex] is not equivalent to [tex]\(-2\pi\)[/tex].
- So, [tex]\( x = -2\pi \)[/tex] is not an asymptote.
2. [tex]\( x = -\frac{\pi}{6} \)[/tex]
- For [tex]\( x = -\frac{\pi}{6} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(-\frac{\pi}{6}\)[/tex] cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = -\frac{\pi}{6} \)[/tex] is not an asymptote.
3. [tex]\( x = \pi \)[/tex]
- For [tex]\( x = \pi \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\pi = \frac{2\pi}{2}\)[/tex], which cannot be expressed as [tex]\(\frac{(2n+1)\pi}{2}\)[/tex].
- So, [tex]\( x = \pi \)[/tex] is not an asymptote.
4. [tex]\( x = \frac{3\pi}{2} \)[/tex]
- For [tex]\( x = \frac{3\pi}{2} \)[/tex], we need to check whether it can be written in the form [tex]\( \frac{(2n+1)\pi}{2} \)[/tex].
- [tex]\(\frac{3\pi}{2} = \frac{(2 \cdot 1 + 1)\pi}{2}\)[/tex], which matches the form [tex]\(\frac{(2n+1)\pi}{2}\)[/tex] for [tex]\( n = 1 \)[/tex].
- Therefore, [tex]\( x = \frac{3\pi}{2} \)[/tex] is an asymptote.
Therefore, the correct answer is:
[tex]\[ x = \frac{3\pi}{2} \][/tex]
The option that corresponds to this is:
[tex]\[ 4 \][/tex]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.