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steel11

  • 2 years ago

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

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

  2. steel11
    • 2 years ago
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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.

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

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

  5. steel11
    • 2 years ago
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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

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

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

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

  9. klimenkov
    • 2 years ago
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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.

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

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

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

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

  14. klimenkov
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

  23. steel11
    • 2 years ago
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    :o?

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

  25. klimenkov
    • 2 years ago
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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?

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

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

  28. steel11
    • 2 years ago
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    and 0

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

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

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

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

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

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

  35. steel11
    • 2 years ago
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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?

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

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

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

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

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

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

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

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

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

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

  46. klimenkov
    • 2 years ago
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    None?

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

  48. steel11
    • 2 years ago
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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.

  49. klimenkov
    • 2 years ago
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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.

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