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anonymous

  • one year ago

if, for all x, f'(x)=(x-2)^4(x-1)^3 , it follows that the function f has

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  1. myininaya
    • one year ago
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    so you need help integrating then \[\int\limits (x-2)^2(x-1)^3 dx \] Let u=x-1 then du=dx if u=x-1 subtracting one on both sides gives: u-1=x-2 so you have \[\int\limits_{}^{}(u-1)^2u^3 du\] this is just to make the multiplication less harsh we still have to multiply

  2. myininaya
    • one year ago
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    so multiply out (u-1)^2 then multiply each one of those terms by u^3 then integrate term by term

  3. myininaya
    • one year ago
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    oh you aren't asking to find f

  4. myininaya
    • one year ago
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    what are you asking for

  5. anonymous
    • one year ago
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    i just started learning calculus and came across this problem on a practice test could u tell me specifically what this is called so i can search it up and learn it

  6. myininaya
    • one year ago
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    what is the question exactly though ?

  7. anonymous
    • one year ago
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    i was asking for the relative minimum

  8. anonymous
    • one year ago
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    i just dont kno how to approach this

  9. myininaya
    • one year ago
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    well you can find the critical numbers of f by setting f'=0 and also finding where f' dne (we don't have to worry about this case here because f' is a polynomial) f'=0 means we need to solve (x-2)^2(x-1)^3=0 again these x's that satisfy (x-2)^2(x-1)^3=0 are critical numbers

  10. myininaya
    • one year ago
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    the critical numbers is where you can possibly have relative mins or maxes

  11. anonymous
    • one year ago
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    oh so i just set x-1=0

  12. myininaya
    • one year ago
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    and x-2=0

  13. myininaya
    • one year ago
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    so you have critical numbers x=1 or x=2 now it is possible to have a rel max or min or neither at the critical numbers

  14. anonymous
    • one year ago
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    oh kk and 1 and 2 are the critical number but the relative minimum is 1 since that is the lowest

  15. myininaya
    • one year ago
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    I would rather not find f'' we can use f' and draw a number line in test intervals around these numbers to see if f is decreasing or increasing

  16. myininaya
    • one year ago
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    |dw:1439324950653:dw|

  17. myininaya
    • one year ago
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    well we don't know that

  18. myininaya
    • one year ago
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    i mean yes 1 is less than 2 but that doesn't mean f(1) is a relative min

  19. myininaya
    • one year ago
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    |dw:1439325010596:dw|

  20. myininaya
    • one year ago
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    I chose a random number from each of the intervals

  21. myininaya
    • one year ago
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    that were divided by the critical numbers x=1 and x=2

  22. myininaya
    • one year ago
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    can you find f'(0) and f'(1/2) and f'(4)?

  23. myininaya
    • one year ago
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    you actually only need the sign not the numerical value

  24. myininaya
    • one year ago
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    for example f'(0)=(-2)^4(-1)^3 all we need to know is that f'(0) is negative which means f is decreasing before we get to x=1

  25. anonymous
    • one year ago
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    f'(0)=- f'(.5)=- f'(4)=+

  26. myininaya
    • one year ago
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    |dw:1439325170086:dw|

  27. myininaya
    • one year ago
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    now we want to know what is happening between 1 and 2 so we need to find the sign of f'(1/2)

  28. myininaya
    • one year ago
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    oh let me check didn't see you post that sorry

  29. myininaya
    • one year ago
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    |dw:1439325249002:dw|

  30. myininaya
    • one year ago
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    so we have no relative max

  31. myininaya
    • one year ago
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    we only have a relative min

  32. myininaya
    • one year ago
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    at that occurs at x=2 but the rel min value is given by f(2)

  33. myininaya
    • one year ago
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    f(1) was neither a rel max or a rel min because the function did not have a switch from decreasing to increasing or the other way around

  34. anonymous
    • one year ago
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    so rel min is at x=2?

  35. myininaya
    • one year ago
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    yep

  36. anonymous
    • one year ago
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    thats also the absolute min right?

  37. myininaya
    • one year ago
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    |dw:1439325403944:dw| because on this interval we can see the lowest value of is f(2)

  38. anonymous
    • one year ago
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    oh lol we made a mistake

  39. myininaya
    • one year ago
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    what is the mistake

  40. anonymous
    • one year ago
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    its 1.5

  41. myininaya
    • one year ago
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    lol yes

  42. anonymous
    • one year ago
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    so 1 is minimum

  43. myininaya
    • one year ago
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    that was dumb sorry

  44. anonymous
    • one year ago
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    no its okay it helped me learn thnks

  45. myininaya
    • one year ago
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    |dw:1439325632222:dw|

  46. myininaya
    • one year ago
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    so we have to fix somethings

  47. anonymous
    • one year ago
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    i unerstand it now its k

  48. myininaya
    • one year ago
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    |dw:1439325670045:dw|

  49. myininaya
    • one year ago
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    so the graph looks something like this and no I don't know if f(2) is 0 this is just a very very rough drawing

  50. myininaya
    • one year ago
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    so yes you are right f(1) is rel min

  51. myininaya
    • one year ago
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    and f(2) is neither rel min or max

  52. myininaya
    • one year ago
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    and from the graph it does appear f(1) is also a abs min

  53. myininaya
    • one year ago
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    you were asking for examples or a site for which this is called: http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx

  54. anonymous
    • one year ago
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    appreciate it

  55. myininaya
    • one year ago
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    np

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