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Math Help - Antiderivative question

  1. #1
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    Antiderivative question


    im pretty sure about everything but the last one..
    i guessed it but for some reason i have a good feeling about my guess.

    could you please explain it to me?
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  2. #2
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    The last one is correct.

    But why is the first one false ? Remember e is just a constant.

    \frac{d}{dx}(x^5+e)=5x^4

    For the second one: if particle has constant acceleration a then
    v(t)=\int adt=at+C_1

    And

    x(t)=\int v(t)=\int at+C_1=\frac{1}{2}at^2+C_1t+C_2

    So it is not a cubic polynomial but a square polynomial.

    For the fifth one:

    Let F is an antiderivative of f and G is antiderivative of g. Then (using the product rule):

    (FG)'=F'G+FG'

    And that is not fg.
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  3. #3
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    Quote Originally Posted by ganur View Post

    im pretty sure about everything but the last one..
    i guessed it but for some reason i have a good feeling about my guess.

    could you please explain it to me?

    1 is true (just derivate x^5 =e to check)
    2 is false (the position would be a quadratic polynomial)
    3 is false
    4 is true
    5 is false (and big time!)
    6 is false

    Tonio
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  4. #4
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    hmmm, looks like we disagree on nr 6.

    If F and G are antiderivatives of f then F-G=C for some constant C\in\mathbb{R}
    Now since F(2)>G(2). We know that F(2)-G(2)=C>0

    Then it follows that F(4)+C=G(4), and since C>0,
    then F(4)>G(4).

    Is this not correct ? If not can you clarify Tonio ?
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  5. #5
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    Quote Originally Posted by hjortur View Post
    hmmm, looks like we disagree on nr 6.

    If F and G are antiderivatives of f then F-G=C for some constant C\in\mathbb{R}
    Now since F(2)>G(2). We know that F(2)-G(2)=C>0

    Then it follows that F(4)+C=G(4), and since C>0,
    then F(4)>G(4).

    Is this not correct ? If not can you clarify Tonio ?

    You are, of course, right. I made a stupid mistake. Thanx

    Tonio
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  6. #6
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    Thank you, I was beginning to doubt myself.
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  7. #7
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    wow thats great help guys! really appreciate that!
    though, something isnt working out for me..
    for the first one, i had no idea that e is the same as c.. but okay i get it.
    2. i get..
    3.&4. i already had that
    5. that makes sense.. yeah i didnt think of that and now im thinking that is so easy
    6. i understand your argument, and that the answer is 'true'

    so here are the answer i put in:
    T, F, F, T, F, T

    and it doesnt work
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  8. #8
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    The trick is that nr 3 is true.

    Rewrite cos^2(x)=\frac{1}{2}+\frac{1}{2}cos(2x)

    Then you get

    2-cos^2(x)=2-\frac{1}{2}-\frac{1}{2}cos(2x)

    And then take the derivative, remember the chain rule.

    \frac{d}{dx}\left(2-\frac{1}{2}-\frac{1}{2}cos(2x)\right)=\frac{1}{2}sin(2x)\cdot2  =sin(2x).

    I am 100% sure everything other is correct.

    Hope that helps.
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