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Help with Calculus please?

Discussion in 'BlackHat Lounge' started by Designerdude13, Mar 11, 2012.

  1. Designerdude13

    Designerdude13 Power Member

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    Hey,

    I know most of you guys have degrees in some type of engineering or science and have taken calculus before. SOOOO can someone help me with this problem lol?? I got the first part right, but not the second

    http://i39.tinypic.com/2cfs4sz.png
     
    Last edited: Mar 11, 2012
  2. kitteh101

    kitteh101 Regular Member

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    What the hell.......... i looked at that equation and it gave me cancer. I make money online but i do not even remotely dabble in that complex math shyt.
     
  3. masterwaldo

    masterwaldo Registered Member

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    If you asked this 12 years ago, I could probably answer it. :D
     
  4. Designerdude13

    Designerdude13 Power Member

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    I don't want to either lol...this math is dumb. I'm not even going to use Calc. But I don't want to fail either lol
     
  5. flc735

    flc735 Regular Member

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    (2,-6) neither
    (4,2) max
    (10,-9) min

    this is highly likely to be wrong, its been a few years!
     
  6. 2makemoney

    2makemoney Regular Member

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    How can I shrink the picture to make it easier to read?
     
  7. flc735

    flc735 Regular Member

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    edit: tried to draw a graph. failed
     
  8. Designerdude13

    Designerdude13 Power Member

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    Yea those are the answers to the first part lol... just x=2, and x=4 --- not 10,-9 and 2,-6 would be a min

    Thanks though
     
  9. flc735

    flc735 Regular Member

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    ctrl and -
     
  10. ramdom

    ramdom Junior Member

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    damn i forgot this so fast I need to review my shit again my last calculus 2 was 1 year ago and i made an "A"
     
  11. flc735

    flc735 Regular Member

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    is the 2nd part asking you to integrate?
     
  12. artizhay

    artizhay BANNED BANNED

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    There should be no critical points in the table given the table represents a function's derivative (y = f'(x)).
     
  13. Designerdude13

    Designerdude13 Power Member

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    No, we haven't got there yet. It would be much easier to that, but plus they don't give us a function

    Are you sure because I only have 1 attempt left. I thought a critical point for y = f'(x) is when the graph crosses the x - axis
     
  14. artizhay

    artizhay BANNED BANNED

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    That's correct. Are any points on that table, given the table is y = f'(x), on the x-axis? Unless you are allowed to enter decimal values? But it asks for points on the table...
    If you are allowed to enter decimal values, calculate the slope between the points that cross the x-axis and use that to find the x-coordinate of the critical point.
     
  15. ben10023

    ben10023 Regular Member

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    f'(x) is a differential not an integral!

    Just plot it as a graph, then think about how the differential is the gradient of f(x)
     
  16. Designerdude13

    Designerdude13 Power Member

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    Yea we are allowed to use decimals because I tried entering "none" before and it was wrong. So do I use the mean value theorem to calculate the slope between the two points?
     
  17. maestropanda

    maestropanda Junior Member

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    You have to graph the table using the numbers as slopes then look at where the highest and lowest peaks are.

    Assuming that it's a closed interval, then it should be something like this:
    http://screensnapr.com/v/UlLjgu.jpg
     
    Last edited: Mar 11, 2012
  18. ben10023

    ben10023 Regular Member

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  19. artizhay

    artizhay BANNED BANNED

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    Okay that clears that up.

    You would use the slope to calculate the x-value of y = 0. I guess you could use MVT. But you really need to just traverse a % of the distance between the two points using the slope to find the point you're looking for.

    Take a look at the points A(4,2) and B(5,-1). We know there is a critical point between the two. We also know that the critical point is closer to (5,-1) because C(x,0) is closer to (5,-1) than (4,2).

    (x,0) is 67% away from (4,2) because the delta-y between the two points is -3 and we want to traverse -2 to get to (x,0). -2/-3 = 0.67

    Therefore, take your initial x position, 4, and move forward 67% of the delta-x. Making the x-value of (x,0) equal to 4.67.

    The formula for that would be
    C.x = 4 + 0.67(5 - 4)
    C.x = A.x + percentMovement(B.x - A.x)

    You can verify this with
    C.y = 2 + 0.67[1 - (-2)] = -0.01
    or
    C.y = A.y + percentMovement(B.y - A.y)
     
    • Thanks Thanks x 1
    Last edited: Mar 11, 2012
  20. Designerdude13

    Designerdude13 Power Member

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    Yes, that was first part. I got that right! This doesn't apply to the second part.

    I have a calculator - Ti-84 Plus - Its not helping for the second part of the problem though. I plotted the graph already

    I was just saying you can make a function an integral by taking the anti-derivative. But we aren't allowed to do that yet. I plotted it already, but it doesn't help with the second part of the problem. And wth is a gradient LOL? I'm thinking of photoshop lol