anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • 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
  • anonymous
@dpaInc
anonymous
  • anonymous
i'm confused. what are you trying to do?

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anonymous
  • anonymous
derivatives are correct...
anonymous
  • anonymous
Application of differentiation Graphing
anonymous
  • anonymous
oh, I see. graph f using f' and f''. is that right?
anonymous
  • anonymous
yeah exactly!!!
anonymous
  • 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
  • anonymous
ok let's start with f first. domain of f is any x except x=0... you'll have an asymptote there.
anonymous
  • anonymous
are there any zeros of f?
anonymous
  • anonymous
yeah positive in both left and right
anonymous
  • anonymous
my horizontal asympto is 0 y=0
anonymous
  • anonymous
correct for horizontal asymptote. but do you have any zero's? (yes, horizontal asymptotes CAN be crossed)
anonymous
  • anonymous
as in? u mean my intercepts? or my relative(local)extrema?
anonymous
  • anonymous
zeros = x-intercept(s)
anonymous
  • anonymous
(-1,0)
anonymous
  • anonymous
good. this is what we have so far... |dw:1332641878342:dw|
anonymous
  • anonymous
now lets tackle the relative max/min.
anonymous
  • anonymous
wait! my y intercept is infinity,is that ok?
anonymous
  • 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
  • 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
  • anonymous
ok so using\[f \prime(x)=-2(x+2)/x^3,\]
anonymous
  • anonymous
got my relative minimum as (-2,-1/2)
anonymous
  • anonymous
my cn"s (-2,0) because the make both the numerator zero
anonymous
  • anonymous
and denominator
anonymous
  • anonymous
|dw:1332641921535:dw|
anonymous
  • anonymous
critical number x=-2 is correct. there is only 1. agree?
anonymous
  • anonymous
so the denominator is not necessary in this case?
anonymous
  • anonymous
not necessary because at we don't have a y value at x = 0 remember? there's an asymptote there.
anonymous
  • anonymous
so i should clean it up,and dont include it on that number line above?
anonymous
  • anonymous
|dw:1332642777320:dw| since we only have one critical number at x = -2 |dw:1332642880662:dw|
anonymous
  • anonymous
with your analysis, your basically done!
anonymous
  • anonymous
u have forgotten\[f \prime \prime\] which will have hyper cn's as well
anonymous
  • 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
  • anonymous
you want point of inflection also? that's what the second derivative is for.
anonymous
  • anonymous
ok, set the second derivative to zero what do you get?
anonymous
  • anonymous
my point of inflation is (-3,-4/3)
anonymous
  • anonymous
i got f(-3) = -4/9
anonymous
  • anonymous
you get the x coordinate from the f'' but stick that x value into the function to get the y coordinate
anonymous
  • anonymous
so inflection point should be (-3, -4/9)
anonymous
  • anonymous
my proffesor plugged it into f(x),in his examples
anonymous
  • anonymous
yep, that's what i did...
anonymous
  • anonymous
oh! lol u are right,forgot to square the 3 below.am sorry
anonymous
  • anonymous
so... |dw:1332643499776:dw| now you're done!
anonymous
  • 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
  • anonymous
your interval check for concavity is (-inf, -3), (-3, 0), (0, inf)
anonymous
  • anonymous
|dw:1332643173272:dw|
anonymous
  • anonymous
got this
anonymous
  • anonymous
nice job!!!
anonymous
  • anonymous
how does that relate to the graph? dont really understand that part
anonymous
  • anonymous
|dw:1332643771538:dw|
anonymous
  • anonymous
as per your analysis...
anonymous
  • anonymous
wow!!! nice,so where is the vertical asymptoe!? zero or -2?
anonymous
  • anonymous
the vertical asymptote is x=0 (the y-axis), your graph cannot cross this line.
anonymous
  • anonymous
so i should not draw tthe dotted lines u drew at -2?
anonymous
  • anonymous
no, not really it's just for clarity that you have a min there.
anonymous
  • 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
  • anonymous
incase of exam
anonymous
  • anonymous
it's whatever x-value makes the denominator of the original function 0.
anonymous
  • anonymous
that is the cn"s only of the original function?
anonymous
  • 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
  • anonymous
if i o that,my V.a will be 1
anonymous
  • anonymous
2x+2/x^2=0 =1
anonymous
  • anonymous
set your denominator which is x^2 equal to zero. x^2 = 0. solve. x = 0. x=0 is you asymptote.
anonymous
  • anonymous
am beginning to think the cn of the 1st derivative is the V.A
anonymous
  • anonymous
no. you do not need derivatives to find vertical asymptotes.
anonymous
  • anonymous
ok what if u have a function like this f(x)=2x/(x+3)^2
anonymous
  • anonymous
then your denomitor is (x+3)^2. so set that to 0: (x + 3)^2 = 0 x = -3 is your vertical asymptote.
anonymous
  • anonymous
ok,dont even need to factor,just have to move the 3 to the right?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=graph+f%28x%29%3D2*x%2F%28x%2B3%29%5E2+with+asymptote
anonymous
  • anonymous
well the denominator is already factored.
anonymous
  • 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
  • anonymous
sorry, I need to leave now. hope this was productive for you as it was for me...
anonymous
  • anonymous
yes... to that last question.. and good work..
anonymous
  • anonymous
yeah! sure it was,thanks alot for your help.atleast i ow understand the topic well
anonymous
  • anonymous
were being kicked out of the lab!!!

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