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Minimum and Maximum Values

  1. Feb 24, 2017 #1
    1. The problem statement, all variables and given/known data

    The problem is in the attached file. The part I need a little help with is part b.

    2. Relevant equations and attempt at a solution

    For part a, I got h(8) = 2, h'(6) = -2, and h''(4) = -2.
    For part c, I found that the integral from 0 to 5 is 7, so I multiplied 7 by 7 to get 49 (because 35 = 5*7). And f(5) = 0, so f(108) also equals 0. I got the tangent line y - 49 = 0(x - 35).

    For part b, I think we're supposed to consider x = 0 and x = 3, since this is where the graph starts increasing from decreasing and vice versa. But I also think we have to evaluate the integral at these points. Can someone explain this?
     

    Attached Files:

  2. jcsd
  3. Feb 24, 2017 #2

    haruspex

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    The period is not 5, so why is that interesting?
     
  4. Feb 24, 2017 #3

    BvU

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    If I consider
    as an attempt, I agree with 0 and 3 but I don't see it decrease before 0... I also see another interval where f is not bigger than at 0 ...

    If you want help for part (a) -- and I think you need it -- , please post your working in detail.

    [edit] strike because the question is about ##h## and I mistakenly read ##f## --- bedtime :sleep:
     
  5. Feb 24, 2017 #4

    haruspex

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    Did you make a similar mistake here? I agree with all the answers for a).
     
  6. Feb 24, 2017 #5

    BvU

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    YES !! aaaarrghh
     
  7. Feb 24, 2017 #6
    I thought the graph would keep repeating every 5 units, but that's not true.
    It seems to repeat every 8 units, so could we do this?

    integral from 0 to 8 = 1
    8*4 = 32
    1*4 = 4
    integral from 0 to 3 = 3
    4 + 3 = 7
    y - 7 = 0(x - 35)?
     
  8. Feb 24, 2017 #7

    haruspex

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    You are told it does.
    Didn"t you calculate it as 2 in part a?
    I do not understand that step. What is the gradient of h at x=35?
     
  9. Feb 25, 2017 #8
    I didn't count the triangle from 0 to 1 when I first calculated the integral from 0 to 8. I think it should be 1 because the area of the trapezoid is -6, area of the large triangle is 8, and area of the small triangle is -1. So 8 + (-6) + (-1) = 1.

    For the derivative at x = 35...
    h(35) = integral from 0 to 35 = 7
    h'(35) = f(35) = 4 because the graph repeats every 8 units...f(3) = 4 for the remaining 3 units after 8*4 = 32. So the derivative at x = 35 is 4.
    If I put it into point-slope form, y - 7 = 4(x - 35) is the tangent line.
     
  10. Feb 25, 2017 #9

    haruspex

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    The trapezoid extends to x=9.
     
  11. Feb 25, 2017 #10
    Ugh. I'm making a lot of mistakes here.

    You're right, we only need the area of the trapezoid up to x = 8. So then the area of the trapezoid is 5, and h(8) = 2. The integral from 0 to 35 is 2*4 + 3 = 11.
    y - 11 = 4(x - 35)
     
  12. Feb 25, 2017 #11

    haruspex

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    Looks right.
     
  13. Feb 25, 2017 #12
    How do I go about part b?
     
  14. Feb 25, 2017 #13

    haruspex

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    Do you understand about local extrema and absolute extrema?
    How do you find a local extremum?
     
  15. Feb 26, 2017 #14
    The absolute maximum is the highest point over the entire graph. The absolute minimum is the lowest point over the graph. The relative max/min is the maximum or minimum over a certain interval.

    I think the first step is to find do h'(x), which is just f(x) and find the critical points and endpoints of the graph. The critical points are 1 and 5, and the endpoints are 0 and 7. And then we need to evaluate g at those values. I came up with the following:
    h(0) = 0
    h(1) = -1
    h(5) = 7
    h(7) = 4
    The absolute min is -1 and max is 7. The relative min is 0 and max is 4.
     
  16. Feb 26, 2017 #15

    haruspex

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    You are asked for the absolute min and max over the whole interval, so yes, -1 and 7.
     
  17. Feb 26, 2017 #16
    Okay. Thank you very much for the help!
     
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