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]

- TrojanPoem

- chestercat

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- anonymous

start by taking the first and second derivatives

- TrojanPoem

I know the steps, the problem lays in this example specifically. No turning points although there is upwards and downwards concavity.

- anonymous

there is a vertical asymptote where denominator is equal to 0

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## More answers

- TrojanPoem

Clarify a bit .

- TrojanPoem

You mean something like this ? http://hotmath.com/hotmath_help/topics/rational-functions/image008.gif

- TrojanPoem

So there is no turning points as the curve is separated into two parts ?

- anonymous

https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example

- anonymous

\[f(x)=\frac{x^2+9}{x}=x+\frac{9}{x}\] right?

- TrojanPoem

yep.

- TrojanPoem

I will watch the video while answering

- 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

i always graph these problems first to get an idea of what im looking to identify

- 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

Good way of doing it, x = 0 -8, 0 > dec 0 , 8 increase 8 = infinite

- TrojanPoem

From now , I will graph it too.

- TrojanPoem

WHere is the turning point satellite ?

- 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

turning point satellite ?

- anonymous

if "turning point" means where it goes from increasing to decreasing and vice versa, those are the zeros of the derivative

- TrojanPoem

Fine, so x= ?

- 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

@billj5 , Finding the asymptote takes decade !

- TrojanPoem

This is the first derivative , we need the second one. and x = 0

- anonymous

lol its not too bad once you get the hang of it

- TrojanPoem

At least , I understood why x = 0 was not a turning point :D , Thanks !

- TrojanPoem

Hey , may you help me with another question ? Will open new one.

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