Question

How do you find the domain and range of this:
y =3csc(3x+π)-2

Answer 1

think I got it... domain is (-infinity, +infinity) Range: [-5,1]

Answer 2

csc(x) = 1/sin(x) so we know we will have a domain problem when 3x+pi = 0, pi, 2pi, ....correct?

Answer 3

so do you think the domain is all real numbers?

Answer 4

@201kbotz ?

Answer 5

arghh... darn okay now I see...

haha oops Thanks zzrock3r

Answer 6

so x is not equal to π+/- 2πn right??

Answer 7

oh but the phase shift would then make it X is not equal to 1/3 +/- 2Pn

Answer 8

*2πn

Answer 9

I for the range wrong too.. I just totally did the sine version of the problem

So the actual range would be (-infinity, -5] U [1, + infinity)

Answer 10

x != ...-2pi,-pi,0,pi,2pi...

Answer 11

oh wait why?

Answer 12

1/sin(x) if x = ...-2pi,-pi,0,pi,2pi... is undefined

Answer 13

oh I see

Answer 14

but that doesn't include the phase shift right?

Answer 15

thats all you have to worry about as far as domain.

To find range take the inverse function, the domain of the inverse is the range of the function

Answer 16

o yeah sorry

Answer 17

so 3x+pi != ...-2pi,-pi,0,pi,2pi...

Answer 18

Okay I see.. thanks.. did I get the range correct:
(-oo, -5] U [1, +oo)

Answer 19

3x+pi != pi*k for any integer k

so

x != (pi*k-pi)/3 or x != (pi/3)(k-1)

Answer 20

oh so that's how you find the domain... I always wondered how people got the k multiple thing... XD

Answer 21

wait on clarification if I had a problem like tan(theta)
then would the domain would be:

X != π(+/-)k

Answer 22

Thanks soo much!!

Answer 23

tan(x) = sin(x)/cos(x)

note: our domain problems come in to play when we are dealing with 1/x, sqrt(x), ln(x), and a few other things...

so our problem with tan(x) is one like 1/x where we dont want to devide by 0. So, with tan(x) we are concered with 1/cos(x). cos(x) != 0. Thus != ?