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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
    • one year ago
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    the first derivative of your function is: \[\Large f'\left( x \right) = \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}}\]

  2. Michele_Laino
    • one year ago
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    am I right?

  3. anonymous
    • one year ago
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    yes

  4. Michele_Laino
    • one year ago
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    ok! Now we have to establish where that first derivative is positive

  5. Michele_Laino
    • one year ago
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    So we have to solve this inequality: \[\Large \frac{{ - 2x}}{{{{\left( {{x^2} - 9} \right)}^2}}} > 0\]

  6. Michele_Laino
    • one year ago
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    what are the solutions of that inequality?

  7. anonymous
    • one year ago
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    would you set the denominator to = 0? i got +-3

  8. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Oh okay that makes sense

  11. Michele_Laino
    • one year ago
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    ok! Now what are the solutions of: \[\Large - 2x > 0\]

  12. anonymous
    • one year ago
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    it would be 0

  13. Michele_Laino
    • one year ago
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    if I multiply, both sides of that inequality by -1, I get: \[\Large 2x < 0\] am I right?

  14. anonymous
    • one year ago
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    yes

  15. Michele_Laino
    • one year ago
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    ok! Now please, divide both sides of the last inequality by 2, what do you get?

  16. anonymous
    • one year ago
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    x<0 (0/2)

  17. Michele_Laino
    • one year ago
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    perfect!

  18. Michele_Laino
    • one year ago
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    that means our function is an incresing function when x<0 and a decreasing function when x>0: |dw:1435207357424:dw|

  19. Michele_Laino
    • one year ago
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    increasing*

  20. Michele_Laino
    • one year ago
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    so what can you conclude?

  21. anonymous
    • one year ago
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    Okay, would my intervals be then increasing from (-infinity,0) and decreasingfrom (0,-infinity)?

  22. Michele_Laino
    • one year ago
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    that's right! Please keep in mind that at subsequent points: x=-3 and x=3, our function is not defined

  23. Michele_Laino
    • one year ago
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    your last statement means that x=0 is a point of local maximum for f(x), am I right?

  24. Michele_Laino
    • one year ago
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    decreasing in (0, +infinity) and increasing in (-infinity, 0)

  25. anonymous
    • one year ago
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    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
    • one year ago
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    no, they are not. Those points don't belong to the domain of our function

  27. anonymous
    • one year ago
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    i thought both infinities were (-)?

  28. anonymous
    • one year ago
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    oh okay!

  29. Michele_Laino
    • one year ago
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    at x=3 and x=-3 we have two vertical asymptotes

  30. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yes! that's right, we have an increasing function in: (-infinity, 0) and a decreasing function in: (0, +infinity)

  33. anonymous
    • one year ago
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    oh i just got it, i kept looking at the y values but its the x values that matter

  34. anonymous
    • one year ago
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    so for my local max, it would just be x=0?

  35. Michele_Laino
    • one year ago
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    yes! That's right!

  36. anonymous
    • one year ago
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    and the function doesn't have a local min right?

  37. Michele_Laino
    • one year ago
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    that's right, we have an unique point of local maximum for f(x)

  38. anonymous
    • one year ago
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    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
    • one year ago
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    here we can compute the second derivative of f(x)

  40. anonymous
    • one year ago
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    6(x^2+3)/(x^2-9)^3

  41. Michele_Laino
    • one year ago
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    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
    • one year ago
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    the same result! :)

  43. Michele_Laino
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    now, since 1>0, then f ''(x)>0 when: \[\Large {x^2} - 9 > 0\] am I right?

  47. anonymous
    • one year ago
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    yes

  48. anonymous
    • one year ago
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    what about the power of 2 though?

  49. Michele_Laino
    • one year ago
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    ok! what are the solutions of: \[\Large {x^2} - 9 > 0\]?

  50. anonymous
    • one year ago
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    you would get +- 3

  51. Michele_Laino
    • one year ago
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    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
    • one year ago
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    oh okay!

  53. Michele_Laino
    • one year ago
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    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
    • one year ago
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    yes!

  55. Michele_Laino
    • one year ago
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    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
    • one year ago
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    please remember there is a theorem which states that a function is concave up in all points for which f ''(x) >0

  57. anonymous
    • one year ago
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    okay! thank you so much!!

  58. Michele_Laino
    • one year ago
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    :)

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