TrojanPoem one year ago 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]

1. anonymous

start by taking the first and second derivatives

2. TrojanPoem

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

3. anonymous

there is a vertical asymptote where denominator is equal to 0

4. TrojanPoem

Clarify a bit .

5. TrojanPoem

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

6. TrojanPoem

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

7. anonymous
8. anonymous

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

9. TrojanPoem

yep.

10. TrojanPoem

I will watch the video while answering

11. 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

12. anonymous

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

13. anonymous

as for the absolute max on $$[1,6]$$ check the critical point (where $$f'(x)=0$$ and also check $$f(1)$$ and $$f(6)$$

14. TrojanPoem

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

15. TrojanPoem

From now , I will graph it too.

16. TrojanPoem

WHere is the turning point satellite ?

17. 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

18. TrojanPoem

turning point satellite ?

19. anonymous

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

20. TrojanPoem

Fine, so x= ?

21. 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

22. TrojanPoem

@billj5 , Finding the asymptote takes decade !

23. TrojanPoem

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

24. anonymous

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

25. TrojanPoem

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

26. TrojanPoem

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