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steel11 Group Title

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

  • 2 years ago
  • 2 years ago

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

    • 2 years ago
  2. steel11 Group Title
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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.

    • 2 years ago
  3. steel11 Group Title
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    I even put the function into a graphing calculator. i still cant get the right answer.

    • 2 years ago
  4. klimenkov Group Title
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    Can you type here what you've got?

    • 2 years ago
  5. steel11 Group Title
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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

    • 2 years ago
  6. klimenkov Group Title
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    Ok. I will type an answer for you.

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

    • 2 years ago
  8. steel11 Group Title
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    all those answers are right. just need 7 and 8

    • 2 years ago
  9. klimenkov Group Title
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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.

    • 2 years ago
  10. klimenkov Group Title
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    Oh. Sorry. I must check it again!

    • 2 years ago
  11. klimenkov Group Title
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    Sorry, I'm not very good in English. Am I right?|dw:1351890529218:dw|

    • 2 years ago
  12. steel11 Group Title
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    left is concave downward, right is concave up. so you have it backwards

    • 2 years ago
  13. steel11 Group Title
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    |dw:1351890663750:dw|

    • 2 years ago
  14. klimenkov Group Title
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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.

    • 2 years ago
  15. klimenkov Group Title
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    I can solve it right here if you want. It will be not so long.

    • 2 years ago
  16. steel11 Group Title
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    your right. no need to go any further. :P

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

    • 2 years ago
  18. steel11 Group Title
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    ye, i was having a tough time with that. :l

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

    • 2 years ago
  20. klimenkov Group Title
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    Ok, ask anything you like.

    • 2 years ago
  21. steel11 Group Title
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    http://gyazo.com/3d89eca12903c12ca28e608233e9f102 find critical points

    • 2 years ago
  22. klimenkov Group Title
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    Hate this congratulations :)

    • 2 years ago
  23. steel11 Group Title
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    :o?

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

    • 2 years ago
  25. klimenkov Group Title
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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?

    • 2 years ago
  26. steel11 Group Title
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    3, and -3

    • 2 years ago
  27. klimenkov Group Title
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    Yes, they are critical.

    • 2 years ago
  28. steel11 Group Title
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    and 0

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

    • 2 years ago
  30. klimenkov Group Title
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    Are you sure that 0 is a critical point?

    • 2 years ago
  31. steel11 Group Title
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    your right, its not. f'(0) does not equal 0

    • 2 years ago
  32. steel11 Group Title
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    but it wont even accept -3 or 3

    • 2 years ago
  33. steel11 Group Title
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    ugh. wait. i think f'(0)= 0. so it must be a critical point

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

    • 2 years ago
  35. steel11 Group Title
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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?

    • 2 years ago
  36. klimenkov Group Title
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    In points -3 and 3 it doesn't exist, so both of them are critical.

    • 2 years ago
  37. steel11 Group Title
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    ok so that is the answer?

    • 2 years ago
  38. klimenkov Group Title
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    Yes. What is the definition of the critical point on your site?

    • 2 years ago
  39. steel11 Group Title
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    http://gyazo.com/a531cae84cdcc8b853ca750f1e2b5e0c

    • 2 years ago
  40. steel11 Group Title
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    im not entirely sure. '_'

    • 2 years ago
  41. klimenkov Group Title
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    Try to type -3,3 insted of 3,-3

    • 2 years ago
  42. steel11 Group Title
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    http://gyazo.com/0354e3dc53f2b3172ff9cbbfd156f2a1

    • 2 years ago
  43. steel11 Group Title
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    this is the problem im having. it wont accept anything.

    • 2 years ago
  44. klimenkov Group Title
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    Hm..did you type None into this field?

    • 2 years ago
  45. steel11 Group Title
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    omg. that actually worked

    • 2 years ago
  46. klimenkov Group Title
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    None?

    • 2 years ago
  47. steel11 Group Title
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    yes. Thanks a lot for the help =]

    • 2 years ago
  48. steel11 Group Title
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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.

    • 2 years ago
  49. klimenkov Group Title
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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.

    • 2 years ago
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