hello need help,plotting differentiattion graph,but am totally confuse in this case as my cn's are {-2,0)
and my hyper cn's {-3,0).how do i do this in the graph?

- anonymous

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- anonymous

This are my given equation\[f(x)=2(x+1)/x^2 , f \prime(x)=-2(x+2)/x^3 and f \prime \prime(x)=4(x+3)/x^4\]

- anonymous

@dpaInc

- anonymous

i'm confused. what are you trying to do?

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## More answers

- anonymous

derivatives are correct...

- anonymous

Application of differentiation
Graphing

- anonymous

oh, I see. graph f using f' and f''. is that right?

- anonymous

yeah exactly!!!

- anonymous

have taken some steps already,can given u my answers if u like,but cant actually do the graphing cos my asnwers look confusing to me

- anonymous

ok let's start with f first. domain of f is any x except x=0... you'll have an asymptote there.

- anonymous

are there any zeros of f?

- anonymous

yeah positive in both left and right

- anonymous

my horizontal asympto is 0
y=0

- anonymous

correct for horizontal asymptote.
but do you have any zero's? (yes, horizontal asymptotes CAN be crossed)

- anonymous

as in? u mean my intercepts? or my relative(local)extrema?

- anonymous

zeros = x-intercept(s)

- anonymous

(-1,0)

- anonymous

good. this is what we have so far...
|dw:1332641878342:dw|

- anonymous

now lets tackle the relative max/min.

- anonymous

wait! my y intercept is infinity,is that ok?

- anonymous

remember, at x=0 we have a vertical asymptote. so your graph cannot cross this line. and yes, I gues you could say that when your y-int is infinity you have an asmptote.

- anonymous

but stay away from that... just look at the denominator of your function. whatever makes your denominator = 0 will be your vertical asymptote.

- anonymous

ok so using\[f \prime(x)=-2(x+2)/x^3,\]

- anonymous

got my relative minimum as (-2,-1/2)

- anonymous

my cn"s (-2,0) because the make both the numerator zero

- anonymous

and denominator

- anonymous

|dw:1332641921535:dw|

- anonymous

critical number x=-2 is correct. there is only 1. agree?

- anonymous

so the denominator is not necessary in this case?

- anonymous

not necessary because at we don't have a y value at x = 0 remember? there's an asymptote there.

- anonymous

so i should clean it up,and dont include it on that number line above?

- anonymous

|dw:1332642777320:dw|
since we only have one critical number at x = -2
|dw:1332642880662:dw|

- anonymous

with your analysis, your basically done!

- anonymous

u have forgotten\[f \prime \prime\]
which will have hyper cn's as well

- anonymous

your relative min is (-2, -1/2), the x-axis is a horizontal asymptote, y-axis is a vertical asymptote:
|dw:1332643042980:dw|

- anonymous

you want point of inflection also? that's what the second derivative is for.

- anonymous

ok, set the second derivative to zero what do you get?

- anonymous

my point of inflation is (-3,-4/3)

- anonymous

i got f(-3) = -4/9

- anonymous

you get the x coordinate from the f'' but stick that x value into the function to get the y coordinate

- anonymous

so inflection point should be (-3, -4/9)

- anonymous

my proffesor plugged it into f(x),in his examples

- anonymous

yep, that's what i did...

- anonymous

oh! lol u are right,forgot to square the 3 below.am sorry

- anonymous

so...
|dw:1332643499776:dw|
now you're done!

- anonymous

sorry. you'll have to do the analysis of where it is concave up/down but you can do that the way you did for the increasing/decreasing function.

- anonymous

your interval check for concavity is (-inf, -3), (-3, 0), (0, inf)

- anonymous

|dw:1332643173272:dw|

- anonymous

got this

- anonymous

nice job!!!

- anonymous

how does that relate to the graph? dont really understand that part

- anonymous

|dw:1332643771538:dw|

- anonymous

as per your analysis...

- anonymous

wow!!! nice,so where is the vertical asymptoe!? zero or -2?

- anonymous

the vertical asymptote is x=0 (the y-axis), your graph cannot cross this line.

- anonymous

so i should not draw tthe dotted lines u drew at -2?

- anonymous

no, not really it's just for clarity that you have a min there.

- anonymous

ok thanks alot,one more last thing
how do i know where to pick as my V.A?
is it the number that appears as both the cn and hyper cn?

- anonymous

incase of exam

- anonymous

it's whatever x-value makes the denominator of the original function 0.

- anonymous

that is the cn"s only of the original function?

- anonymous

hmmm. just set the denominator of the original function to zero and solve. that's where your VA will be.
I don't like to use the word cn because your implying derivatives. you do not need derivatives to locate VA.

- anonymous

if i o that,my V.a will be 1

- anonymous

2x+2/x^2=0
=1

- anonymous

set your denominator which is x^2 equal to zero.
x^2 = 0. solve.
x = 0.
x=0 is you asymptote.

- anonymous

am beginning to think the cn of the 1st derivative is the V.A

- anonymous

no. you do not need derivatives to find vertical asymptotes.

- anonymous

ok what if u have a function like this f(x)=2x/(x+3)^2

- anonymous

then your denomitor is (x+3)^2. so set that to 0:
(x + 3)^2 = 0
x = -3 is your vertical asymptote.

- anonymous

ok,dont even need to factor,just have to move the 3 to the right?

- anonymous

http://www.wolframalpha.com/input/?i=graph+f%28x%29%3D2*x%2F%28x%2B3%29%5E2+with+asymptote

- anonymous

well the denominator is already factored.

- anonymous

oh!! oh!! now i get,any number u plog in,into the denominator of your original function that makes it zero is your V.A?

- anonymous

sorry, I need to leave now. hope this was productive for you as it was for me...

- anonymous

yes... to that last question.. and good work..

- anonymous

yeah! sure it was,thanks alot for your help.atleast i ow understand the topic well

- anonymous

were being kicked out of the lab!!!

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