TrojanPoem
  • TrojanPoem
Find the intervals of upwards concavity and downwards concavity and turning points if exists for the curve of function f as f(x) = x^2 +9 /x and find the absolute maximum and minimum for the function when x (- [1, 6]
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
start by taking the first and second derivatives
TrojanPoem
  • TrojanPoem
I know the steps, the problem lays in this example specifically. No turning points although there is upwards and downwards concavity.
anonymous
  • anonymous
there is a vertical asymptote where denominator is equal to 0

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TrojanPoem
  • TrojanPoem
Clarify a bit .
TrojanPoem
  • TrojanPoem
You mean something like this ? http://hotmath.com/hotmath_help/topics/rational-functions/image008.gif
TrojanPoem
  • TrojanPoem
So there is no turning points as the curve is separated into two parts ?
anonymous
  • anonymous
https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example
anonymous
  • anonymous
\[f(x)=\frac{x^2+9}{x}=x+\frac{9}{x}\] right?
TrojanPoem
  • TrojanPoem
yep.
TrojanPoem
  • TrojanPoem
I will watch the video while answering
anonymous
  • anonymous
\[f'(x)=1-\frac{9}{x^2}\]find the interval over what \(f'\) is positive, that will tell you where it \(f\)is increasing
anonymous
  • anonymous
i always graph these problems first to get an idea of what im looking to identify
anonymous
  • anonymous
as for the absolute max on \([1,6]\) check the critical point (where \(f'(x)=0\) and also check \(f(1)\) and \(f(6)\)
TrojanPoem
  • TrojanPoem
Good way of doing it, x = 0 -8, 0 > dec 0 , 8 increase 8 = infinite
TrojanPoem
  • TrojanPoem
From now , I will graph it too.
TrojanPoem
  • TrojanPoem
WHere is the turning point satellite ?
anonymous
  • anonymous
in any case if you want a nice picture and pretty much everything else use this http://www.wolframalpha.com/input/?i=%28x^2%2B9%29%2Fx
TrojanPoem
  • TrojanPoem
turning point satellite ?
anonymous
  • anonymous
if "turning point" means where it goes from increasing to decreasing and vice versa, those are the zeros of the derivative
TrojanPoem
  • TrojanPoem
Fine, so x= ?
anonymous
  • anonymous
idk i didn't do it solve \[1-\frac{9}{x^2}=0\] and you will get them can probably do it in your head
TrojanPoem
  • TrojanPoem
@billj5 , Finding the asymptote takes decade !
TrojanPoem
  • TrojanPoem
This is the first derivative , we need the second one. and x = 0
anonymous
  • anonymous
lol its not too bad once you get the hang of it
TrojanPoem
  • TrojanPoem
At least , I understood why x = 0 was not a turning point :D , Thanks !
TrojanPoem
  • TrojanPoem
Hey , may you help me with another question ? Will open new one.

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