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]

Question

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]

anonymous · Answer

start by taking the first and second derivatives

trojanpoem · Answer

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

anonymous · Answer

there is a vertical asymptote where denominator is equal to 0

trojanpoem · Answer

Clarify a bit .

trojanpoem · Answer

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

trojanpoem · Answer

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

anonymous · Answer

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

anonymous · Answer

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

trojanpoem · Answer

yep.

trojanpoem · Answer

I will watch the video while answering

anonymous · Answer

$$f'(x)=1-\frac{9}{x^2}$$find the interval over what $f'$ is positive, that will tell you where it $f$is increasing

anonymous · Answer

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

anonymous · Answer

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 · Answer

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

trojanpoem · Answer

From now , I will graph it too.

trojanpoem · Answer

WHere is the turning point satellite ?

anonymous · Answer

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 · Answer

turning point satellite ?

anonymous · Answer

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

trojanpoem · Answer

Fine,  so x= ?

anonymous · Answer

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 · Answer

@billj5  , Finding the asymptote takes decade !

trojanpoem · Answer

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

anonymous · Answer

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

trojanpoem · Answer

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

trojanpoem · Answer

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