Question

Is there a way to find the domain and range of y=tan(2x-pi) without a calculator? If yes, how?

Answer 1

memory ...

Answer 2

the range of tan is not effected by an shifts in period or other..

Answer 3

the domain is tho .... recall that the normal domain is between -pi\2 and pi/2

Answer 4

-pi/2 < theta < pi/2

solve it to get x

Answer 5

so the domain would be (-pi/2, pi/2), and range be (-infinity, +infinity)?

Answer 6

here theta is 2x - pi

Answer 7

(2x-pi) ; factor out the 2

2(x- pi/2) ; the domain for tan is then pi/2 ; shifted by pi/2 to the right I believe ...

Answer 8

I'm sorry, I don't really understand. Why would I need to factor out the 2? If it is -pi/2 < 2x - pi < pi/2, then wouldn't I add pi to all, then divide everything by 2 to get:

pi/4 < x< 3pi/4?

Answer 9

it has to do with how a trig function is transformed

Answer 10

you got it denebel the answer !! : )

Answer 11

tan(x) is the normal looking trig function; we modify the period with a "w" such that tan(wx) has a period of pi/w right?

Answer 12

a phase shift is where we move the graph left or right; tan(x-p) would shift it

Answer 13

tan(wx-p) accounts for both of these transformations such that we first factor out the "w" to see our phace shift ... if i recall it correctly

Answer 14

http://msenux.redwoods.edu/wagner/math25/supp/phaseshift/phaseshift.pdf

Answer 15

wolfram gives us this to look at ..

http://www.wolframalpha.com/input/?i=tan%282x-pi%29

Answer 16

amistre64: pi/2 is the period right..?

Ishaan94: The domain will be (pi/4, 3pi/4)? And the range can be (-infinity, +infinity)?

Answer 17

yeah i think so denebel

Answer 18

Thanks :)

wait, amistre64: for the link http://www.wolframalpha.com/input/?i=tan%282x-pi%29, what does the result (tan(2x)) mean?

Answer 19

i was wondering that meself, i beleive it means that modifying it by pi has no real effect to it since the period is pi to begin with

Answer 20

pi/2 is the period yes

Answer 21

this wolfram shows that tan(2x-pi) is the same as tan(2(x- pi/2)).

I tend to not be so confident when working from me memory ;)

http://www.wolframalpha.com/input/?i=y%3Dtan%282x-pi%29+and+y%3Dtan%282%28x-%28pi%2F2%29%29%29

Answer 22

o___o I am a little confused now.

According to Wolfram, the domain when this equation will be 'one to one' is (-pi/4, pi/4)... unless I am reading the graph wrong?? But I calculated it to be (pi/4, 3pi/4)....

Answer 23

if we take the normal span of tan(x) we go from -pi/2 to pi/2 as a "domain"

-pi/2 < 2(x- pi/2) < pi/2 ; divide out the 2

-pi/4 < x- pi/2 < pi/4 ; + 2pi/4 to get.

pi/4 < x < 3pi/4 ... which is really the same as:

-pi/4 < x < pi/4

Answer 24

<....-pi/4..........pi/4..........3pi/4............5pi/4.........> etc ..

Answer 25

Uhm how is it that it is -pi/2<2(x-pi/2) < pi/2? I thought it would 2x-pi, instead of 2(x-pi/2)?

Answer 26

But by Wolfram, it is proved that the two equations are equal..

Answer 27

...how would I know about the second equation if only the first equation is given to me? By memory?

Answer 28

the issue with periodic functions is that they can be expressed in more than one way; which makes them a little tricky to deal with mentally

Answer 29

the proper way to define the domain would be to include/exclude; all the possibilities that arise from -inf to inf

Answer 30

the domain us all x such that x notEqual to some iteration of an add "pi/4".

Answer 31

... its like i had a stroke while typing lol

Answer 32

If I were to include all, wouldn't that not make it a 'one to one' function anymore?

How could I express it? Would it be like (pi/4 + k*pi, 3pi/4 + k*pi)?

Answer 33

{x|x \(\ne n*\cfrac{pi}{4}\)} where n is an odd value .... there is a better way to express it but I cant think of it at the moment :)

to make it a one-to-one function we would have to restrict the domain, and by convention that would be between -pi/4 and pi/4, but convention is just code for, "do as i say" lol