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steel11

Where is this function concave up and concave down. x^2/(x-5)^2

  • one year ago
  • one year ago

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  1. klimenkov
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    \[\frac{x^2}{(x-5)^2}\]Can you take derivatives?

    • one year ago
  2. steel11
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    yes. I know i have to take the second derivative, but i having trouble finding out where the 2nd derivative equals zero.

    • one year ago
  3. steel11
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    I even put the function into a graphing calculator. i still cant get the right answer.

    • one year ago
  4. klimenkov
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    Can you type here what you've got?

    • one year ago
  5. steel11
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    going to take me a while to get it, have to do chain rule and everything. '_' you can just post the answer if you want. >.> already been on this problem for like an hour

    • one year ago
  6. klimenkov
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    Ok. I will type an answer for you.

    • one year ago
  7. steel11
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    this was a multiple part question, this is what i have so far. http://gyazo.com/003eb77ccf9977811aa71cf7f65fb35f

    • one year ago
  8. steel11
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    all those answers are right. just need 7 and 8

    • one year ago
  9. klimenkov
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    \(f''(x)=\frac2{(-5 + x)^2} - \frac{8 x}{(-5 + x)^3} + \frac{6 x^2}{(-5 + x)^4}\) The horizontal asymtote isn't right.

    • one year ago
  10. klimenkov
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    Oh. Sorry. I must check it again!

    • one year ago
  11. klimenkov
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    Sorry, I'm not very good in English. Am I right?|dw:1351890529218:dw|

    • one year ago
  12. steel11
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    left is concave downward, right is concave up. so you have it backwards

    • one year ago
  13. steel11
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    |dw:1351890663750:dw|

    • one year ago
  14. klimenkov
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    Ok. So when \(x\in(-\infty,-\frac52)\) it is concave downward. \(x\in(-\frac52,5)\bigcup(5,+\infty)\) it is concave upward.

    • one year ago
  15. klimenkov
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    I can solve it right here if you want. It will be not so long.

    • one year ago
  16. steel11
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    your right. no need to go any further. :P

    • one year ago
  17. klimenkov
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    All you have to do is to solve this:\[\frac2{(-5 + x)^2} - \frac{8 x}{(-5 + x)^3} + \frac{6 x^2}{(-5 + x)^4}=0\]

    • one year ago
  18. steel11
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    ye, i was having a tough time with that. :l

    • one year ago
  19. steel11
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    hmm. while your here. could you answer 1 small part to another question? pretty straight forward this time

    • one year ago
  20. klimenkov
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    Ok, ask anything you like.

    • one year ago
  21. steel11
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    http://gyazo.com/3d89eca12903c12ca28e608233e9f102 find critical points

    • one year ago
  22. klimenkov
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    Hate this congratulations :)

    • one year ago
  23. steel11
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    :o?

    • one year ago
  24. steel11
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    people told me there are not critical points to that function, but im not entirely sure if their correct.

    • one year ago
  25. klimenkov
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    The definition of the critical point says that it is that point where the derivative of the function is equal to zero or DOES NOT EXIST. It is your case. Wat are the point where it doesn't exist?

    • one year ago
  26. steel11
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    3, and -3

    • one year ago
  27. klimenkov
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    Yes, they are critical.

    • one year ago
  28. steel11
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    and 0

    • one year ago
  29. steel11
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    but my site says 3, -3 and 0 are not critical points http://gyazo.com/77d0af7eb8d1e3ce273ff32c24eb58ca

    • one year ago
  30. klimenkov
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    Are you sure that 0 is a critical point?

    • one year ago
  31. steel11
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    your right, its not. f'(0) does not equal 0

    • one year ago
  32. steel11
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    but it wont even accept -3 or 3

    • one year ago
  33. steel11
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    ugh. wait. i think f'(0)= 0. so it must be a critical point

    • one year ago
  34. klimenkov
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    No. It is not right. The derivative is:\[\frac{-4 ( x^2+9 )}{( x^2-9 )^2}\]

    • one year ago
  35. steel11
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    oh. so what do you do from there. set it equal to zero and solve? that would be -3 and 3 right?

    • one year ago
  36. klimenkov
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    In points -3 and 3 it doesn't exist, so both of them are critical.

    • one year ago
  37. steel11
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    ok so that is the answer?

    • one year ago
  38. klimenkov
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    Yes. What is the definition of the critical point on your site?

    • one year ago
  39. steel11
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    http://gyazo.com/a531cae84cdcc8b853ca750f1e2b5e0c

    • one year ago
  40. steel11
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    im not entirely sure. '_'

    • one year ago
  41. klimenkov
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    Try to type -3,3 insted of 3,-3

    • one year ago
  42. steel11
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    http://gyazo.com/0354e3dc53f2b3172ff9cbbfd156f2a1

    • one year ago
  43. steel11
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    this is the problem im having. it wont accept anything.

    • one year ago
  44. klimenkov
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    Hm..did you type None into this field?

    • one year ago
  45. steel11
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    omg. that actually worked

    • one year ago
  46. klimenkov
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    None?

    • one year ago
  47. steel11
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    yes. Thanks a lot for the help =]

    • one year ago
  48. steel11
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    i did a combined total of 180 tries on both of the question i asked you. finally got the right answers.

    • one year ago
  49. klimenkov
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    It is strange. Ask your teacher the definition of the critical point. Hope you will solve different problems easily in the future. Good luck for you. I wish you a lot of well. Hope you understand.

    • one year ago
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