- anonymous

In the function f(x) = sec 2x, what are the following:
- Domain
- Asymptotes

- chestercat

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- SolomonZelman

Domain will have gaps.
\(f(x)=\sec^2(x)\)
\(f(x)=\displaystyle \frac{1}{ \cos^2(x)}\)
so for every value where cos²(x)=0, you have a vertical asymptote.

- anonymous

Wait, so \[\sec^2x \] is also \[\sec2x\]?

- anonymous

Can I use the T-table to graph the function or nah?

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## More answers

- SolomonZelman

Also, you know that |cos x| < |sec x|
And cos(x) (as well as sine), is between 1 and -1.
This way, |sec x| is always ≥1 and ≤-1.
So, it will look roughly like this:
|dw:1437070613045:dw|

- anonymous

Also, is there any way to find the domain without graphing?

- SolomonZelman

wait, I thought you didn't use a caret (^), but your function is sec(2x)?

- anonymous

Yup it is sec(2x).

- SolomonZelman

So, it will be same as sec²x still. JUST THAT:
the arcs will be thinner.
the asymptote is everywhere when cos(x)=0

- SolomonZelman

But it is still same, because if cos(x)=0, then cos²(x)=0 as well./

- anonymous

Yes, that's why the range is \[(-\infty, -1], [1,\infty)\] correct?

- SolomonZelman

yes

- SolomonZelman

((With a U in the middle ))

- SolomonZelman

(-∞,-1] U [1,∞)

- anonymous

oh yeah right. I thought you said no haha.

- anonymous

So the period is pi.

- anonymous

And there is no amplitude, right?

- SolomonZelman

yes, period is π

- SolomonZelman

yes no amplitude.

- SolomonZelman

(I am lagging, my connection is bad)

- anonymous

So the domain is all real nos. except ±π/4, ±3π/4, ±5π/4

- SolomonZelman

yes\(\color{red}{,}\) no amplitude.
(like that)

- SolomonZelman

yes, and so on.... for all values tof x, that make cos(x)=0

- SolomonZelman

of* x

- anonymous

Also, in the book, it says the domain is all real nos except
\[\frac{ \pi }{ 2}+\pi n\]

- anonymous

What does it mean? Is it the same with what I stated above?

- SolomonZelman

for all integer values of n....
that is the pattern in which these numbers are generated

- SolomonZelman

i need to refresh. sorry. hold on.

- anonymous

Okay, so how do I put the domain like that?

- SolomonZelman

well, you can say that same thing in the book, and don't forget to add that \({n \in {\bf Z}\)

- SolomonZelman

ops

- SolomonZelman

So, you can state (for the asymptotes) that
\(\large\color{black}{ \displaystyle \frac{\pi }{2} +\pi\cdot n;~\left\{n \in {\bf Z} \right\} }\)

- anonymous

|dw:1437071434398:dw|

- SolomonZelman

you can use wolframalpha.com to fill in the table.

- SolomonZelman

it is a very cool calculator.

- anonymous

No, I don't want just the answers.

- SolomonZelman

What do you want then?

- anonymous

I want to learn.

- SolomonZelman

Oh, I mean not a graphing calc. I mean to find particular values of cse(2x), if you want.
Or, you can plug the values yourself, and calculate them yourself. And if you encounter a problem calculating (by hand) one value or the other, you can ask me.

- anonymous

Yup, I want to plug the values myself.

- anonymous

The thing is I don't know how to start. I just know that the "count number" is pi/4.

- anonymous

\[\frac{ 1 }{ 4 } \times (period)\]

- SolomonZelman

don't really know what 1/4 • period means

- anonymous

Okay, so how would I find the x-values then?

- anonymous

THANK YOU FOR THE HUGE HELP @SolomonZelman !

- anonymous

oops didn't mean to use uppercase lol

- anonymous

Domain: all real numbers other than: ±π/4, ±3π/4, ±5π/4...
Range:(-∞,-1) U (1,∞)
Period: π
Asymptotes:x=±π/4, x=±3π/4, x=±5π/4...

- pooja195

yes and?

- pooja195

Can you pm and delete that reply please ?

- pooja195

Im not a mod but i can talk to you about it.

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