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This are my given equation\[f(x)=2(x+1)/x^2 , f \prime(x)=-2(x+2)/x^3 and f \prime \prime(x)=4(x+3)/x^4\]
i'm confused. what are you trying to do?
derivatives are correct...
Application of differentiation Graphing
oh, I see. graph f using f' and f''. is that right?
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
ok let's start with f first. domain of f is any x except x=0... you'll have an asymptote there.
are there any zeros of f?
yeah positive in both left and right
my horizontal asympto is 0 y=0
correct for horizontal asymptote. but do you have any zero's? (yes, horizontal asymptotes CAN be crossed)
as in? u mean my intercepts? or my relative(local)extrema?
zeros = x-intercept(s)
good. this is what we have so far... |dw:1332641878342:dw|
now lets tackle the relative max/min.
wait! my y intercept is infinity,is that ok?
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.
but stay away from that... just look at the denominator of your function. whatever makes your denominator = 0 will be your vertical asymptote.
ok so using\[f \prime(x)=-2(x+2)/x^3,\]
got my relative minimum as (-2,-1/2)
my cn"s (-2,0) because the make both the numerator zero
critical number x=-2 is correct. there is only 1. agree?
so the denominator is not necessary in this case?
not necessary because at we don't have a y value at x = 0 remember? there's an asymptote there.
so i should clean it up,and dont include it on that number line above?
|dw:1332642777320:dw| since we only have one critical number at x = -2 |dw:1332642880662:dw|
with your analysis, your basically done!
u have forgotten\[f \prime \prime\] which will have hyper cn's as well
your relative min is (-2, -1/2), the x-axis is a horizontal asymptote, y-axis is a vertical asymptote: |dw:1332643042980:dw|
you want point of inflection also? that's what the second derivative is for.
ok, set the second derivative to zero what do you get?
my point of inflation is (-3,-4/3)
i got f(-3) = -4/9
you get the x coordinate from the f'' but stick that x value into the function to get the y coordinate
so inflection point should be (-3, -4/9)
my proffesor plugged it into f(x),in his examples
yep, that's what i did...
oh! lol u are right,forgot to square the 3 below.am sorry
so... |dw:1332643499776:dw| now you're done!
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.
your interval check for concavity is (-inf, -3), (-3, 0), (0, inf)
how does that relate to the graph? dont really understand that part
as per your analysis...
wow!!! nice,so where is the vertical asymptoe!? zero or -2?
the vertical asymptote is x=0 (the y-axis), your graph cannot cross this line.
so i should not draw tthe dotted lines u drew at -2?
no, not really it's just for clarity that you have a min there.
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?
incase of exam
it's whatever x-value makes the denominator of the original function 0.
that is the cn"s only of the original function?
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.
if i o that,my V.a will be 1
set your denominator which is x^2 equal to zero. x^2 = 0. solve. x = 0. x=0 is you asymptote.
am beginning to think the cn of the 1st derivative is the V.A
no. you do not need derivatives to find vertical asymptotes.
ok what if u have a function like this f(x)=2x/(x+3)^2
then your denomitor is (x+3)^2. so set that to 0: (x + 3)^2 = 0 x = -3 is your vertical asymptote.
ok,dont even need to factor,just have to move the 3 to the right?
well the denominator is already factored.
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?
sorry, I need to leave now. hope this was productive for you as it was for me...
yes... to that last question.. and good work..
yeah! sure it was,thanks alot for your help.atleast i ow understand the topic well
were being kicked out of the lab!!!