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FreeTrader

  • 3 years ago

Factoring Advice: Goal is to eliminate the x's. [(x+y)^2 - x^2] / y

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  1. fiddlearound
    • 3 years ago
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    it's equal to 2x+y , eliminate the x's ?

  2. fiddlearound
    • 3 years ago
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    oh that's different than the first one

  3. FreeTrader
    • 3 years ago
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    I think there is a way to do it, but I am not sure how I can get them to all cancel out.

  4. FreeTrader
    • 3 years ago
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    \[((x+y)2−y2)/y \] goal is to eliminate the x's and be left with only y's to deal with...

  5. FreeTrader
    • 3 years ago
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    Thoughts? I have tried the usual things I can think of, but unless I made an error I am not catching, I am stumped.

  6. FreeTrader
    • 3 years ago
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    BRB

  7. fiddlearound
    • 3 years ago
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    is this part of a LIM problem ?

  8. FreeTrader
    • 3 years ago
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    Yes, how on earth did you guess?

  9. fiddlearound
    • 3 years ago
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    I mean are you looking for: \[\lim_{y \rightarrow 0} \frac{(x+y)^2-x^2}{y}\]

  10. FreeTrader
    • 3 years ago
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    Uh, indeed I am...

  11. FreeTrader
    • 3 years ago
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    Can you help me get unstuck? I really want to learn how to do this...

  12. FreeTrader
    • 3 years ago
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    I have it down to y + 2x and I don't know if I can use x in my lim answer, I've never seen a variable in a lim.

  13. fiddlearound
    • 3 years ago
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    \[\lim_{y \rightarrow 0} \frac{\cancel{x^2}+2xy+y^2-\cancel{x^2}}{y}=\lim_{y \rightarrow 0}2x+y=2x\]

  14. FreeTrader
    • 3 years ago
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    Is it acceptable to have a variable in the lim solution?

  15. fiddlearound
    • 3 years ago
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    yes, coz it's basically the derivative: \[(x^2)'=2x\]

  16. FreeTrader
    • 3 years ago
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    Or is this DNE?!?

  17. fiddlearound
    • 3 years ago
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    the final answer is 2x which is the derivative of x^2

  18. FreeTrader
    • 3 years ago
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    I don't know enough to understand that in the context of this problem. I understand what a first derivative is, and how you showed that, but I do not understand it in relation to my first "evaluate lim's" homework assignment.

  19. fiddlearound
    • 3 years ago
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    how was the problem worded ?

  20. FreeTrader
    • 3 years ago
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    So the lim is really 2x?!?

  21. FreeTrader
    • 3 years ago
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    The lim stmt you wrote, preceded by the word "Evaluate:"

  22. fiddlearound
    • 3 years ago
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    yes, so the answer would be 2x

  23. FreeTrader
    • 3 years ago
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    We haven't covered derivatives yet - not until tomorrow. So this is kinda tricky!

  24. fiddlearound
    • 3 years ago
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    this is how they introduce you to derivatives

  25. fiddlearound
    • 3 years ago
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    you don't have to know what a derivative is to solve the LIM problem. But in effect, they will reveal that what you were doing was calculating the derivative.

  26. FreeTrader
    • 3 years ago
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    Have you seen Gilbert Strang's high school calculus course on MIT OCW, called "The Big Picture of Calculus?" THAT's how to teach calculus.

  27. FreeTrader
    • 3 years ago
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    Are you a student or faculty?

  28. fiddlearound
    • 3 years ago
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    that formula is basically calculating rise over run (where they make the run smaller and smaller)

  29. FreeTrader
    • 3 years ago
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    Ahhhh

  30. fiddlearound
    • 3 years ago
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    neither. I studied math about 20 years ago.

  31. FreeTrader
    • 3 years ago
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    What do you do now that allows you to stay fresh on it? I am 44, BTW.

  32. fiddlearound
    • 3 years ago
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    we are very close in age :) I just started solving problems on openstudy to keep the mind from rusting too much.

  33. FreeTrader
    • 3 years ago
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    Well, thank you for your help. I am now a fan.

  34. fiddlearound
    • 3 years ago
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    BTW, I started going over the MIT OCW single variable calculus - but then got sucked in over here in the math group.

  35. FreeTrader
    • 3 years ago
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    I want to build up a community on MIT OCW SCHOLAR SVC.

  36. FreeTrader
    • 3 years ago
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    I am trying to login there every night at 8 PM. I missed it tonight, the first time others were there, too.

  37. FreeTrader
    • 3 years ago
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    So, if you are around at 8PM ET, I will try to be there in the MIT OCW SVC (NOT the 18.01 - the SCHOLAR version. It is a new and improved course, totally self contained.

  38. FreeTrader
    • 3 years ago
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    Thanks. Over and out on this case. I had solved it and didn't realize I was done.

  39. fiddlearound
    • 3 years ago
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    My schedule in the evenings is not fixed, juggling around kids & work . And I didn't continue past the first OCW lecture - btw I didn't know they now have the new and improved course.

  40. FreeTrader
    • 3 years ago
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    It is wonderful. I hope you find time to give it a try during your "me time."

  41. FreeTrader
    • 3 years ago
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    I'm going to resume my homework. I have about a zillion more to go...

  42. fiddlearound
    • 3 years ago
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    OK :)

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