Question

need to find horizontal asymptote of
(sin x)/( 1+ cos x) using limit when x->infinity

Answer 1

1+cosx=0
when x=pie rad= 180 degree

Answer 2

:/ im confused

Answer 3

i think that would be a vertical asymptote, not a horizontal

Answer 4

:/

Answer 5

wow, so it turns out that
\[\frac{\sin(x)}{1+\cos(x)} = \tan(\frac{x}{2}) \]
tan doesnt have any horizontal asymptotes...so i guess this function doesnt either?

Answer 6

okay, thanks! :)

Answer 7

gj joe i didn't see it was =tan(x/2)

Answer 8

it would have been alot easier if i had seen that

Answer 9

just so everyone knows
veritcal asymptotes or holes occur when the function does not exist
horizontal asymptotes are those thingys you find when you let x approach infinity

Answer 10

"those thingys" lol

Answer 11

it is my mathematical notation
it is my mathematical variable

Answer 12

(not that i can think of a better name at this time >.>)

Answer 13

:)

Answer 14

i could not sleep
its 6:49 am here

Answer 15

go into the thread by aross, help me solve that 10x 10 matrix lol

Answer 16

10 by 10 thats crazy
do that using an algorithm on a computer

Answer 17

i put it in my phone, its actually solvable =/

Answer 18

ok i will look at it

Answer 19

another question...finding critical points of (e^-x) (sin x)

Answer 20

because I know that in division, I just have to look for the first derivative, and the value that make the numerator = 0, but not the denominator, is the critical point...but in this case, which one is the critical point?

Answer 21

f'(x)=-e^(-x)(sinx)+e^(-x)*cosx
f'(x)=e^(-x){-sinx+cosx)=0
e^(-x) never 0
so we have -sinx+cosx=0
sinx=cosx
sin and cos are the same when x=pi/4
and when x=5pi/4
so we can generate a thingy for all the critical numbers
x=2npi+pi/4
and
x=2npi+5pi/4
for n=...,-3,-2,-1,0,1,2,3,...

Answer 22

sounds good to me.

Answer 23

i added 2npi
because we can keep going around the circle over and over again until we get dizzy and then some more

Answer 24

the sun is coming up <.<

Answer 25

the sun is up

Answer 26

where is the 10 by 10 matrix?

Answer 27

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e12f0780b8b56e555994ac7

Answer 28

so the critical point of (e^-x)(sin x) are 2n pi+pi/4 and 2n pi + 5pi/4

Answer 29

and since you say critical point
we should say
(2npi+pi/4,f(2npi+pi/4))
and
(2npi+5pi/4,f(2npi+5pi/4))

Answer 30

i named your function f by the way i hope that was okay

Answer 31

it's okay ;)

Answer 32

:)

Answer 33

myininaya, I have to find the intervals in which f(x) is increasing or decreasing....i read that I have to look for the zeros of the first derivative of f(x)...what else do I have to do after looking for the zeroes of f'(x)??....what f'(x) doesn't have zeroes??
....how do I find the increasing or decreasing intervals??

Answer 34

for example in (x^2 -16) / (x^2 -4x)

Answer 35

the first derivative is -4(x^2) but that doesn't have zeroes or roots...how do I know the increasing and decreasing intervals?

Answer 36

ok f tells where the number is
f' tells us if the function is increasing, decreasing, or neither
f'' tells us if the function is concave up or concave down
so we need f' to find intervals of increase and decrease and where it is neither(when the slope is 0)
so if
\[f(x)=\frac{x^2-16}{x^2-4x}=\frac{(x-4)(x+4)}{x(x-4)}=\frac{x+4}{x},x \neq 4=1+4x^{-1}, x \neq 4\]
\[f'(x)=-4x^{-1-1}=-4x^{-2}=\frac{-4}{x^2}\]
ok we x=0 is a critical number

-----0----
(-inf,0) (0,inf)

we choose test numbers before and after each critical number
so we choose a number before 0 (doesn't matter what number)
f'(-3)=a negative number => the function is decreasing on (-inf,0)
now we choose a number after 0 (doesn't matter what number)
f'(5)=a negative number => the function is also decreasing on (0,inf)

Answer 37

my definition of critical number is any numbers that make f' not exist and also satisfy
f'=0

Answer 38

we do not have the latter here but the first part we do

Answer 39

okay...so, a critical point or critical number is a number that makes the equation = 0?

Answer 40

derivative of equation=0

Answer 41

f'(x) = 0
or
f'(x) = does not exist

this are critical numbers?

Answer 42

some people might also add a little bit to the defintion of the critical number and say:
any number in the domain of f that makes f' not exist or f'=0 is a critical number

Answer 43

yes!

Answer 44

i do not use the domain part very much
like for instance we said above f' does not exist when x=0 above
so x=0 is a critical number
some people might say x=0 is not a critical number because 0 is not in the domain of f
so they might just choose any number from -inf to inf to see if the function is decreasing or increasing
and in fact we see that f'<0 for all numbers except at x=0 (since we have discontinuity;
you can also say it isn't doing anything at x=4 since f doesn't exist there)

Answer 45

okay

Answer 46

let me write this down...this is key for me

Answer 47

just remember critical numbers are numbers satisfying
f' does not exist
and
f'(x)=0
and you will be golden!

Answer 48

i said and
lets say or

Answer 49

perfect!

Answer 50

well you should be golden on finding where f is increasing/decreasing/neither

Answer 51

of course there is alot more cal to learn and not all of it is about critical numbers

Answer 52

you're right!

Answer 53

inflection points are values that make the second derivative = 0 or not exist?

Answer 54

yes! :)

Answer 55

well let me add two more things
it must also satisfy that the concavity switches at that number to be inflection point
it also must be in the domain of the function to be an inflection point

Answer 56

it is possible to not have inflection points??..like here (x^2 -16) / (x^2 -4x)

Answer 57

if a value of the critic point isn't in the domain of the function, then that value isn't a critic point?

Answer 58

or that only applies for inflection points?

Answer 59

f''(x)=8/(x^3)
f'' dne at x=0
but 0 is not in the domain of f (so there is no inflection point)
we can choose test number around 0 to plug into f'' to see when f is concave up/down
---0---
f''(-1)=negative number => concave down
f''(1)=positive number => concave up
critical numbers i would say do not have to be in the domain of f
but
critical points do

Answer 60

inflection points also have to be in the domain to be considered an inflection point

Answer 61

okay

Answer 62

hi again myininaya!

Answer 63

I need to graph few equations, if I don't know how to graph, I can choose random x values and substitute them in the equation?