Question

Graph the following function: y=csc(1/2x) find the domain, range, period, even/odd function, and asymptotes

Answer 1

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

Answer 2

is this
\[\csc(\frac{1}{2x})\] or
\[csc(\frac{1}{2}x)\]?

Answer 3

the 2nd one

Answer 4

can you help?

Answer 5

yeah i hope

Answer 6

\[\csc(\frac{x}{2})=\frac{1}{\sin(\frac{x}{2})}\]

Answer 7

so range is from
\[-\infty\] to \[\infty\]

Answer 8

sine is odd which means
\[\sin(-x)=-\sin(x)\] and therefore this one is odd too

Answer 9

of course you cannot divide by zero, therefore we have to make sure that
\[\sin(\frac{x}{2})\neq0\]

Answer 10

\[sin(x)=0\] means
\[x=...-3\pi,-2\pi,-\pi,0, \pi, 2\pi, 3\pi...\]

Answer 11

we have
\[sin(\frac{x}{2})=0\] giving
\[\frac{x}{2}=k\pi\] and therefore
\[x=2k\pi\] so domain is all real numbers except
\[x = 2k\pi, k\in Z\]

Answer 12

ok i made a mistake earlier when i said range was all real numbers. sine is bounded above by 1 and below by -1 so this function is never between -1 and 1. in other words the domain is
\[(-\infty,-1)\cup (1,\infty)\]

Answer 13

how is it so far?

Answer 14

except of course for my mistake!

Answer 15

I am taking notes as we go, thank you so much

Answer 16

well don't write the range is all real numbers. delete that from your notes!

Answer 17

ok thank you
so range is ?

Answer 18

\[(-\infty,-1]\cup [1,\infty)\]

Answer 19

oh excuse me, i wrote that as the domain

Answer 20

it is -1 when
\[sin(\frac{x}{2})=-1\]and 1 when \[\sin(\frac{x}{2})=1\]

Answer 21

otherwise
\[-1<\sin(\frac{x}{2})<1\]
so
\[\frac{1}{\sin(\frac{x}{2})}>1\] or
\[\frac{1}{\sin(\frac{x}{2})}<-1\]

Answer 22

what is left? asymptotes. we have found those. the function is undefined at
\[x=2k\pi, k\in Z\] and so those are the asymptotes

Answer 23

and finally the period! we know that the period of sine is \[2\pi\] so the period of
\[\frac{1}{\sin(x)}\] is also
\[2\pi\] and therefore the period of
\[\frac{1}{sin(\frac{x}{2})}\] is
\[4\pi\]

Answer 24

and that takes care of all the questions yes?

Answer 25

unless of course you would like to graph it and see what it looks like

Answer 26

http://www.wolframalpha.com/input/?i=y%3Dcsc%28x%2F2%29

Answer 27

yes, thank you so much, i have one problem left if you have timeo look at it

Answer 28

go ahead,

Answer 29

y= tan(\[\pi\]x)+1

Answer 30

that is all suppose to be in a line

Answer 31

that is all suppose to be in a line

Answer 32

\[tan(\pi x)+1\]

Answer 33

?

Answer 34

ok this one is much easier

Answer 35

tangent is odd, but this is not odd because of the 1 outside

Answer 36

in other words if
\[f(x)=\tan(x)+1\] then
\[f(-x)=\tan(-x)+1=-\tan(x)+1\neq -f(x)\]

Answer 37

this one does have range all real numbers because tangent has range all real numbers.

Answer 38

domain we find as before;
\[tan(\pi x)+1=\frac{sin(\pi x)}{cos(\pi x)}+1\] and since you cannot divide by 0 we have to make sure that
\[\cos(\pi x)\neq 0\]

Answer 39

\[\cos(x)=0\] means
\[x=...-\frac{5\pi}{2},-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, ...\]

Answer 40

so
\[cos(\pi x)=0 \] means
\[x=-\frac{5}{2}, -\frac{3}{2}, -\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2}...\]

Answer 41

more succinctly written as
\[x\neq \frac{2k+1}{2}, k\in Z\]

Answer 42

those are you asymptotes as well

Answer 43

and finally the period of tangent is
\[\pi\] so the period of
\[\tan(\pi x)\] is \[\frac{\pi}{\pi}=1\]

Answer 44

and that is that. except for the graph which is here
http://www.wolframalpha.com/input/?i=tan%28pi+x%29%2B1

Answer 45

is this even or neither

Answer 46

no not even or odd.

Answer 47

tangent is odd, but the +1 at the end throws it off

Answer 48

and domain is

Answer 49

all real numbers except
\[x=\frac{2k+1}{2}, k\in Z\]

Answer 50

in other words all numbers except 1/2,3/2,5/2,...

Answer 51

odd integers divided by 2

Answer 52

ok and is there a way to look at the graph

Answer 53

i sent it i think

Answer 54

http://www.wolframalpha.com/input/?i=tan%28pi+x%29%2B1

Answer 55

oh im soryy mine did not show a graph

Answer 56

those vertical lines are just asymptotes not part of the graph. did ou see the second one?

Answer 57

yes thank you
is ther an email adress that i can contact you at when i need additional help

Answer 58

you can try me at satellite73.openstudy@gmail.com

Answer 59

thank you alot i really appreciate it