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Math Help - Limit definitions

  1. #1
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    Limit definitions

    Use to find

    f prime (2) where f(x)=x^2 -3x + 1

    Here's what I've done. Correct my mistakes please.

    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1)/Δx
    lim Δx->0 ((2+Δx)^2-3(2+Δx)+1)/Δx
    lim Δx->0 (-1+Δx+Δx^2)/Δx
    lim Δx->0 -1+Δx^2

    and now i am stuck.
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  2. #2
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    Quote Originally Posted by lsp2010 View Post
    Use to find

    f prime (2) where f(x)=x^2 -3x + 1

    Here's what I've done. Correct my mistakes please.

    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1)/Δx
    lim Δx->0 ((2+Δx)^2-3(2+Δx)+1)/Δx
    lim Δx->0 (-1+Δx+Δx^2)/Δx
    lim Δx->0 -1+Δx^2

    and now i am stuck.
    You've forgotten to subtract f(a).
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  3. #3
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    Can you explain that for me? Or show me? I'm confused.
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  4. #4
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    First try it yourself. Surely you know how to subtract?
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  5. #5
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    Yes, I do. But I thought I already did it. I don't see where I messed up.

    Or should it be

    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1-(x^2-3x+1)/Δx
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  6. #6
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    Quote Originally Posted by lsp2010 View Post
    Use to find

    f prime (2) where f(x)=x^2 -3x + 1

    Here's what I've done. Correct my mistakes please.

    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1)/Δx
    lim Δx->0 ((2+Δx)^2-3(2+Δx)+1)/Δx
    lim Δx->0 (-1+Δx+Δx^2)/Δx
    lim Δx->0 -1+Δx^2

    and now i am stuck.
    Until you get good at these (and I'm sure you will!) I'd suggest that you pull it apart.
    Firstly work out f(2 + deltax) which is what you have attempted to do above.
    Then work out f(2)
    Then subtract: f(2+delta x) - f(2)
    Then put it over delta x

    Do that first then come back.
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  7. #7
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    when should i plug in the 2? at the beginning or end?
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  8. #8
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    Quote Originally Posted by lsp2010 View Post
    when should i plug in the 2? at the beginning or end?
    It doesn't really matter. If you only need to find f prime (2) then put the 2 in at the beginning.
    If you had to find f prime (3) , then f prime (4) etc, you'd leave the a (or c) in until the end, so you don't end up doing the same process for different numbers.
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  9. #9
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    Quote Originally Posted by Debsta View Post
    It doesn't really matter. If you only need to find f prime (2) then put the 2 in at the beginning.
    If you had to find f prime (3) , then f prime (4) etc, you'd leave the a (or c) in until the end, so you don't end up doing the same process for different numbers.
    BTW you were on the right track before, but you didn't subtract the f(2) before dividing by delta x. That's what the other person was trying to tell you.
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  10. #10
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    Quote Originally Posted by Debsta View Post
    It doesn't really matter. If you only need to find f prime (2) then put the 2 in at the beginning.
    If you had to find f prime (3) , then f prime (4) etc, you'd leave the a (or c) in until the end, so you don't end up doing the same process for different numbers.
    Quote Originally Posted by Debsta View Post
    BTW you were on the right track before, but you didn't subtract the f(2) before dividing by delta x. That's what the other person was trying to tell you.
    got to go soon - are you still there?
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  11. #11
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    so when i did subtract it, would it have been
    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1-(x^2-3x+1)/Δx
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  12. #12
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    Quote Originally Posted by lsp2010 View Post
    so when i did subtract it, would it have been
    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1-(x^2-3x+1)/Δx
    Please pull it apart - makes it easier to understand. I'm going to use h instead of delta x to make typing easier.

    f(2+h) = (2+h)^2 - 3 (2+h) +1 = 4 + 4h + h^2 - 6 - 3h +1 = h^2 + h -1

    You had that before.

    Now f(2) = 2^2 - 3x2 + 1 = -1
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  13. #13
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    so when i did subtract it, would it have been
    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1-(x^2-3x+1)/Δx ?
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  14. #14
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    Quote Originally Posted by Debsta View Post
    Please pull it apart - makes it easier to understand. I'm going to use h instead of delta x to make typing easier.

    f(2+h) = (2+h)^2 - 3 (2+h) +1 = 4 + 4h + h^2 - 6 - 3h +1 = h^2 + h -1

    You had that before.

    Now f(2) = 2^2 - 3x2 + 1 = -1

    So f(2+h)-f(2) = ??
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  15. #15
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    Quote Originally Posted by lsp2010 View Post
    so when i did subtract it, would it have been
    lim Δx->0 ((c+Δx)^2-3(c+Δx)+1-(x^2-3x+1)/Δx ?
    Not quite - you should have c (or 2) where you have x in the last bracket.
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