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zeesbrat3

  • one year ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f of x equals 9 times x raised to the one third power plus 9 halves times x raised to the four thirds power . You must justify your answer using an analysis of f ′(x) and f ′′(x).

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  1. zeesbrat3
    • one year ago
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    \[f(x) = 9x^\frac{ 1 }{ 3 } + \frac{ 9 }{ 2 }x^\frac{ 4 }{ 3 }\]

  2. zeesbrat3
    • one year ago
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    I have that \[f'(x) = 3x^\frac{ 1 }{ 3 }[x ^{-2} + 2]\]

  3. zeesbrat3
    • one year ago
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    @ganeshie8

  4. zeesbrat3
    • one year ago
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    So, I set f'(x) = 0 and I got \[x = 0, \pm \frac{ 1 }{ \sqrt{-2} }\]

  5. ganeshie8
    • one year ago
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    http://www.wolframalpha.com/input/?i=%289x%5E%281%2F3%29+%2B+9%2F2*x%5E%284%2F3%29%29%27

  6. zeesbrat3
    • one year ago
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    I'm confused...

  7. zeesbrat3
    • one year ago
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    @jim_thompson5910 @amistre64

  8. zeesbrat3
    • one year ago
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    Hi, Jim_thomspon5910

  9. jim_thompson5910
    • one year ago
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    \[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\] \[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\] \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] \[\Large f \ '(x) = 3x^{1/3}\left(x^{-2} + 2\right)\] so I'm getting the same f ' (x) as you did

  10. zeesbrat3
    • one year ago
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    Then set it to 0, yes?

  11. jim_thompson5910
    • one year ago
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    if you set f ' (x) equal to zero, and solve for x, you do get \[\Large x = 0 \text{ or } x = \pm \frac{1}{\sqrt{-2}}\] so I agree there as well

  12. jim_thompson5910
    • one year ago
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    The portion \(\Large \pm\frac{1}{\sqrt{-2}}\) aren't real numbers, so we can ignore these two solutions

  13. jim_thompson5910
    • one year ago
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    Now either do a first derivative test or a second derivative test to see if a local min or max is at x = 0

  14. zeesbrat3
    • one year ago
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    The number line check?

  15. jim_thompson5910
    • one year ago
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    yeah you can do that for a first derivative test

  16. jim_thompson5910
    • one year ago
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    see if f ' (x) changes sign for a value less than 0, and for a value greater than 0

  17. zeesbrat3
    • one year ago
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    when it is less than 0, it is decreasing, when it is greater than 0, it is increasing

  18. jim_thompson5910
    • one year ago
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    correct on both

  19. zeesbrat3
    • one year ago
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    So 0 is a min

  20. jim_thompson5910
    • one year ago
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    when x < 0, f ' (x) < 0 when x > 0, f ' (x) > 0 we have a local min at x = 0 |dw:1437349622635:dw|

  21. jim_thompson5910
    • one year ago
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    the min is at x = 0 the actual min value itself is unknown at this point

  22. jim_thompson5910
    • one year ago
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    plug x = 0 back into f(x) to find the min value

  23. jim_thompson5910
    • one year ago
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    I'm not thinking if x = 0, then f ' (x) is undefined because of the x^(-2) portion. That leads to a division by zero error. So x = 0 cannot be a critical value. The graph of f(x) also shows a vertical tangent at x = 0

  24. zeesbrat3
    • one year ago
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    Why plug 0 back into f(x)?

  25. ganeshie8
    • one year ago
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    \[\Large f(x) = 9x^{1/3} + \frac{9}{2}x^{4/3}\] \[\Large f \ '(x) = 9(1/3)x^{1/3-1} + \frac{9}{2}*\frac{4}{3}x^{4/3-1}\] \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] \[\Large f \ '(x) = 3x^{1/3}\left(x^{-\color{red}{1}} + 2\right)\]

  26. jim_thompson5910
    • one year ago
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    oh my bad, I factored incorrectly

  27. zeesbrat3
    • one year ago
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    So would the second zero be a critical number?

  28. jim_thompson5910
    • one year ago
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    x = 0 won't work solve x^(-1) + 2 = 0 for x to get the actual only critical value that will work

  29. zeesbrat3
    • one year ago
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    So x = -1/2

  30. jim_thompson5910
    • one year ago
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    yep, do another first derivative test to see if that is a local min, local max, or neither

  31. zeesbrat3
    • one year ago
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    Using 0 as well, or not?

  32. jim_thompson5910
    • one year ago
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    x = -1/2 is the critical value now

  33. jim_thompson5910
    • one year ago
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    you cannot use x = 0 because f ' (x) is undefined when x = 0

  34. zeesbrat3
    • one year ago
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    Would it be a min again?

  35. jim_thompson5910
    • one year ago
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    yes

  36. zeesbrat3
    • one year ago
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    So, when x = -1/2 there is a local minimum

  37. jim_thompson5910
    • one year ago
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    it says `Find the x-coordinates of any relative extrema ` they don't want the y coordinates, so you can stop for this part

  38. jim_thompson5910
    • one year ago
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    yeah

  39. zeesbrat3
    • one year ago
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    Can you help me find the points of inflection please?

  40. jim_thompson5910
    • one year ago
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    you'll need to find f '' (x)

  41. jim_thompson5910
    • one year ago
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    so differentiate f ' (x)

  42. jim_thompson5910
    • one year ago
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    \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] will be easier to work with

  43. zeesbrat3
    • one year ago
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    I got \[f''(x) = \frac{ 2(x-1) }{ x^\frac{ 5 }{ 3 } }\]

  44. jim_thompson5910
    • one year ago
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    I'm getting \[\Large f \ '(x) = 3x^{-2/3} + 6x^{1/3}\] \[\Large f \ ''(x) = 3(-2/3)x^{-2/3-1} + 6(1/3)x^{1/3-1}\] \[\Large f \ ''(x) = -2x^{-5/3} + 2x^{-2/3}\]

  45. jim_thompson5910
    • one year ago
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    You can factor out -2x^(-5/3) \[\Large f \ ''(x) = -2x^{-5/3} + 2x^{-2/3}\] \[\Large f \ ''(x) = -2x^{-5/3}(1 - x)\]

  46. jim_thompson5910
    • one year ago
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    so it looks like we got the same things, just in slightly different forms

  47. zeesbrat3
    • one year ago
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    Yay!

  48. jim_thompson5910
    • one year ago
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    now solve f '' (x) = 0 for x

  49. jim_thompson5910
    • one year ago
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    that will give potential x coordinates of the point of inflection

  50. jim_thompson5910
    • one year ago
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    to see if they actually are at the inflection point, you need to see if f '' changes sign

  51. zeesbrat3
    • one year ago
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    So \[0 = 2(x-1)\]

  52. zeesbrat3
    • one year ago
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    @jim_thompson5910

  53. zeesbrat3
    • one year ago
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    So x = 1 is the point of inflection?

  54. jim_thompson5910
    • one year ago
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    it's at the possible point of inflection

  55. jim_thompson5910
    • one year ago
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    set up a number line then see if f '' changes sign for a value to the left of x = 1 and to the right of x = 1

  56. jim_thompson5910
    • one year ago
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    don't use x = 0

  57. zeesbrat3
    • one year ago
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    I used -1 and 2... From decreasing to increasing I got

  58. jim_thompson5910
    • one year ago
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    when x = 0.5, f '' (x) = -3.1748 when x = 2, f '' (x) = 0.62996 f '' (x) changes from negative to positive as we move through x = 1 so we definitely have an inflection point at x = 1

  59. zeesbrat3
    • one year ago
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    xD

  60. jim_thompson5910
    • one year ago
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    it turns out x = 0 is also at an inflection point. I'm reading now that if f '' (c) doesn't exist, then x = c could be a possible inflection point (assuming f(c) is defined and f '' changes sign) http://www.sosmath.com/calculus/diff/der15/der15.html since f '' (x) changes in sign going from say x = -0.5 to x = 0.5, this means that another inflection point is at x = 0

  61. zeesbrat3
    • one year ago
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    Jeez, math is complicated

  62. jim_thompson5910
    • one year ago
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  63. jim_thompson5910
    • one year ago
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  64. zeesbrat3
    • one year ago
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    Thank you!

  65. jim_thompson5910
    • one year ago
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    no problem

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