Question

domain: tan2x

Answer 1

\[-\Pi \to \Pi\]

Answer 2

no, you can put 2pi in there as well.

Answer 3

what?

Answer 4

The point here is that tan(x) is undefined in 1/2pi.

Answer 5

how do i find the domain lol

Answer 6

ok, the domain is the set of all values you can put in the function.

Answer 7

i know what it is lol, how do i find the domain of y = tan2x

Answer 8

As I said, you can't put 1/2 pi in the function.

Answer 9

so what do i write as an answer....

Answer 10

Well, first note that tan(x) is periodic with period pi, so you can't 3/2pi in there either, or 13/2pi etc.

Answer 11

But it's 2x...

Answer 12

is it tan(x)^2 or tan(2x), I thought the first one.

Answer 13

secondd

Answer 14

He's basically right, though, so pi/4 instead of of pi/2..

Answer 15

lol what is the answerrrr.....

Answer 16

The answer is tan (2x) is not defined for \[\Pi\]/4 + pi/2*k

Answer 17

that's right.

Answer 18

Where's my frickin' cookie!? :p

Answer 19

So domain is everything except that...

Answer 20

except 2/4 + pi/2 * k?

Answer 21

pi/4 + pi/2*k

Answer 22

oookay if you say so lol. what's k?

Answer 23

Constant

Answer 24

k = 1,2,3....I guess.

Answer 25

or -1,-2,-3,...

Answer 26

\[f(x)=\tan(x)=\frac{sin(x)}{cos(x)}\]
the domain of
\[f is (-\frac{\pi}{2},\frac{\pi}{2})\]
so the domain of
\[f(x)=\tan(2x) is (-\frac{\pi}{4},\frac{\pi}{4})\]

Answer 27

i disagree.

Answer 28

Me too...

Answer 29

That isn't the domain of tan (x), that's where it is undefined...

Answer 30

well that is the domain of tan(2x) if we want it to be 1 to 1

Answer 31

and that is the domain of tan(x) if we want it to be 1 to 1

Answer 32

I don't want it to be 1 to 1.

Answer 33

Lol.

Answer 34

if you don't care about it being 1 to 1
then the domain of f=tan(x) is
all real numbers except
\[x=\frac{\pi}{2}+2npi, x=\frac{3\pi}{2}+2npi, n \in \mathbb{Z} \]

Answer 35

Domain of cos is R

Answer 36

none of this makes sense.

Answer 37

We're really bad at explaining this.

Answer 38

haha not even.. i just dont really get it, havent done it in forever, and you guys have conflicting ideas lol

Answer 39

the domain of \[f=\tan(2x)\] is all real numbers except
\[x=\frac{\pi}{4}+npi, x=\frac{3\pi}{4}+npi, n \in \mathbb{Z} \]

Answer 40

idk what all the little symbols are hahah

Answer 41

no, its n* pi/2, the period of tan(x) is pi.

Answer 42

laksjdf;laskejf;laskejf...... so is 1 + sinx neither odd nor even?

Answer 43

i'm done my answer is not changing
i feel confident its right
peace

Answer 44

I feel like saying you're wrong, but I don't know if I should.

Answer 45

I mean, I know I'm right.

Answer 46

lol you can if you want but i'm right

Answer 47

you can't put 1/4 pi +1/2 pi in tan(2x).

Answer 48

what?

Answer 49

that's in your domain.

Answer 50

plus minus (2k+1) pi/4, k in Z

Answer 51

you mean 3pi/4
no 3pi/4 is not in the domain

Answer 52

oh, fuuu

Answer 53

\[f(\frac{3\pi}{4})=\tan(2 \cdot \frac{3\pi}{4})=\tan(\frac{3\pi}{2}) dne\]

Answer 54

thats why we don't include it in our domain

Answer 55

I see, I'm going to blame your notation, I have to blame something.

Answer 56

lol
f me?

Answer 57

you were right, I was wrong.

Answer 58

n is an integer

Answer 59

any integer

Answer 60

:)

Answer 61

I think we all learned a valuable lesson. Everything is okay, as long as you win a medal.

Answer 62

lol

Answer 63

We had the same domain in mind though, I just didn't realize it.

Answer 64

lol

Answer 65

ok :)

Answer 66

I want to hear more about this 1 to 1 business...

Answer 67

its only good if you want the inverse to exist

Answer 68

Ok....

Answer 69

f=tan(x) has vertical asymptotes at pi/2 and -pi/2 right?

Answer 70

Yes, I get it, just o define the reciprocal functions with restricted range.....

Answer 71

i have food

Answer 72

:-)

Answer 73

\[f=\tan(2x)\]
so we don't want tan(2x) to be between
2x=-pi/2 and 2x=pi/2
x=-pi/4 and x=pi/4

Answer 74

oops we want tan(2x) to be between*

Answer 75

(-pi/4,pi/4)

Answer 76

so that f is 1 to 1 and thus the inverse exists

Answer 77

ok i got eat now