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## anonymous one year ago Find the intervals of increase and decrease of 1/x^2-9 b) find the coordinates of the local max and min Please help, I got x=0 and x=+-3 but im not sure what the intervals would be for the increase and decrease

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1. Michele_Laino

the first derivative of your function is: $\Large f'\left( x \right) = \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}}$

2. Michele_Laino

am I right?

3. anonymous

yes

4. Michele_Laino

ok! Now we have to establish where that first derivative is positive

5. Michele_Laino

So we have to solve this inequality: $\Large \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}} > 0$

6. Michele_Laino

what are the solutions of that inequality?

7. anonymous

would you set the denominator to = 0? i got +-3

8. Michele_Laino

no, since the denominator is always positive, then that fraction will be positive, if and only if also the numerator will be positive, namely: $\Large - 2x > 0$

9. Michele_Laino

the denominator: $\Large {{{\left( {{x^2} - 9} \right)}^2}}$ is a square, which is always positive, and it is equal to zero at x=3 and x=-3 That means our first derivative doesn't exist at x=3 and x=-3. Also your function doesn't exist at x=3 and x=-3 for the same reason

10. anonymous

Oh okay that makes sense

11. Michele_Laino

ok! Now what are the solutions of: $\Large - 2x > 0$

12. anonymous

it would be 0

13. Michele_Laino

if I multiply, both sides of that inequality by -1, I get: $\Large 2x < 0$ am I right?

14. anonymous

yes

15. Michele_Laino

ok! Now please, divide both sides of the last inequality by 2, what do you get?

16. anonymous

x<0 (0/2)

17. Michele_Laino

perfect!

18. Michele_Laino

that means our function is an incresing function when x<0 and a decreasing function when x>0: |dw:1435207357424:dw|

19. Michele_Laino

increasing*

20. Michele_Laino

so what can you conclude?

21. anonymous

Okay, would my intervals be then increasing from (-infinity,0) and decreasingfrom (0,-infinity)?

22. Michele_Laino

that's right! Please keep in mind that at subsequent points: x=-3 and x=3, our function is not defined

23. Michele_Laino

your last statement means that x=0 is a point of local maximum for f(x), am I right?

24. Michele_Laino

decreasing in (0, +infinity) and increasing in (-infinity, 0)

25. anonymous

thats what i have, but Im not sure if its right, would the -3 and 3 have anything to do with the local max and min?

26. Michele_Laino

no, they are not. Those points don't belong to the domain of our function

27. anonymous

i thought both infinities were (-)?

28. anonymous

oh okay!

29. Michele_Laino

at x=3 and x=-3 we have two vertical asymptotes

30. anonymous

okay, i see it on the graph, you wrote decreasing (0,+infinity) and increasing (-infinity,0) why is it +infinity on the decreasing part? I thought it was - on both points

31. Michele_Laino

nevertheless those points (x=3, and x=-3) can not be considered as local maximum or minimu points since they don't check the definitions of local maximum or minimum points

32. Michele_Laino

yes! that's right, we have an increasing function in: (-infinity, 0) and a decreasing function in: (0, +infinity)

33. anonymous

oh i just got it, i kept looking at the y values but its the x values that matter

34. anonymous

so for my local max, it would just be x=0?

35. Michele_Laino

yes! That's right!

36. anonymous

and the function doesn't have a local min right?

37. Michele_Laino

that's right, we have an unique point of local maximum for f(x)

38. anonymous

okay! my next question is where its concave up and concave down. for concave up would it be (-3,-infinity)U(3,+infinity)?

39. Michele_Laino

here we can compute the second derivative of f(x)

40. anonymous

6(x^2+3)/(x^2-9)^3

41. Michele_Laino

after a simple computation, I got: $\Large f''\left( x \right) = 6\frac{{{x^2} + 3}}{{{{\left( {{x^2} - 9} \right)}^3}}}$

42. Michele_Laino

the same result! :)

43. Michele_Laino

now the second derivative, can be rewritten as follows: $\Large f''\left( x \right) = \frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}\frac{1}{{\left( {{x^2} - 9} \right)}}$

44. Michele_Laino

and we ahve to establish when f''(x) is positive, namely we have to study this inequality: $\Large f''\left( x \right) > 0 \Rightarrow \frac{1}{{\left( {{x^2} - 9} \right)}} > 0$

45. Michele_Laino

since: $\Large \frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}$ is always positive inside the domain of our function f(x)

46. Michele_Laino

now, since 1>0, then f ''(x)>0 when: $\Large {x^2} - 9 > 0$ am I right?

47. anonymous

yes

48. anonymous

what about the power of 2 though?

49. Michele_Laino

ok! what are the solutions of: $\Large {x^2} - 9 > 0$?

50. anonymous

you would get +- 3

51. Michele_Laino

as you can see I factorized f ''(x) as follows: $\Large f''\left( x \right) = \left[ {\frac{{6\left( {{x^2} + 3} \right)}}{{{{\left( {{x^2} - 9} \right)}^2}}}} \right] \times \frac{1}{{\left( {{x^2} - 9} \right)}}$ now the quantity inside the square brackets is always positive, so we can neglect it when we have to solve the inequality f ''(x)>0

52. anonymous

oh okay!

53. Michele_Laino

now the solution of this inequality: $\Large {x^2} - 9 > 0$ is: $\Large \left( { - \infty , - 3} \right) \cup \left( {3, + \infty } \right)$ right?

54. anonymous

yes!

55. Michele_Laino

so we can conclude that our function is concave up inside that interval, namely in: $\large \left( { - \infty , - 3} \right) \cup \left( {3, + \infty } \right)$ and concave down inside this interval: $\Large \left( { - 3,3} \right)$

56. Michele_Laino

please remember there is a theorem which states that a function is concave up in all points for which f ''(x) >0

57. anonymous

okay! thank you so much!!

58. Michele_Laino

:)

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