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Math Help - tangents with specific slopes...

  1. #1
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    tangents with specific slopes...

    so how do i derive this problem?

    are there any points on the curve y=(x/2)+1/(2x-4) where the slope is -3/2?
    where do i start?

    thx!
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  2. #2
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    Quote Originally Posted by cyberdx16 View Post
    so how do i derive this problem?

    are there any points on the curve y=(x/2)+1/(2x-4) where the slope is -3/2?
    where do i start?

    thx!
    The derivative is,
    y'=(1/2)-2(2x-4)^(-2)
    You need that to be -3/2
    Thus,
    1/2-2(2x-4)^(-2)=-3/2
    Thus,
    -2(2x-4)^(-2)=-2
    Thus,
    (2x-4)^(-2)=1
    Thus,
    2x-4=1
    Thus,
    x=2.5
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  3. #3
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    If you learned MVT here is an alternative.
    ---
    Now there is a longer but more elegant way to show that this is true. It uses the awesome faboulus mean-value theorem (most important theorem in calculus).
    ---
    I am going to find an interval [a,b] such that,
    f(b)-f(a)
    --------
    b-a

    Would be equal to -3/2

    Of course we need to check, if f(x) is continous on [a,b] and differenciable on (a,b).

    Let us use a convienat point say a=0.
    Then,
    f(0)=-1/4

    Thus, the mean value of the interval is,
    f(b)+1/4
    ---------
    b

    We need this to be equal to -3/2
    If we multiply both sides by b we have,
    f(b)+1/4=(-3/2)b
    Thus,
    b/2+1/(2b-4)+1/4=(-3/2)b
    Thus,
    1/(2b-4)+1/4=-2b
    Thus,
    1/(2b-4)+2b=-1/4
    Thus,
    (1+4b^2-8b)/(2b-4)=-1/4
    Thus,
    4b^2-8b+1=1-b/2
    Thus,
    8b^2-16b=-b
    Thus,
    8b^2-15b=0
    Thus,
    b(8b-15)=0
    Thus,
    b=0 makes equation true but we cannot use it since b>a
    Or,
    b=15/8
    Which makes the equation true.
    Keeps the function continous on [0,15/8]
    And differenciable on (0,15/8)

    This mean that, on the interval (0,15/8)
    there exists a point c such that,
    f'(c)=MEAN VALUE=-3/2
    Q.E.D.

    Nice thing about this is that it is an existence theorem which I love.
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  4. #4
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    (2x-4)^(-2)=1
    Thus,
    2x-4=1
    Thus,
    x=2.5[/QUOTE]

    so to get rid of the (2x-4)^(-2)=1
    did you squard both sides to get rid of the ^-2

    EDIT: or i mean from (2x-4)^(-2)=1 to x=2.5 you would multiply by (2x-4)^(-2) which you would get 1=(2x-4)^2 and then sq both sides and set everything to x?
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  5. #5
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    You have,
    (2x-4)^(-2)=1
    Thus,
    Code:
    1
    ----                 =1
    (2x-4)^2
    Thus,
    multiply both sides by (2x-4)^2
    1=(2x-4)^2
    Just take the positive root cuz you are only trying to find one value that makes it true.
    2x-4=1
    -------------
    The mental trick I do it take reciprocal of both sides.
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  6. #6
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    Quote Originally Posted by cyberdx16 View Post
    so how do i derive this problem?

    are there any points on the curve y=(x/2)+1/(2x-4) where the slope is -3/2?
    where do i start?

    thx!
    Slightly generalised:

    You are asked if there are points on the curve y=f(x), where the slope is m

    The slope of the curve at x is dy/dx, so the question becomes does the
    equation:

    dy/dx=m

    have any real solutions.

    In your case f(x)=(x/2)+1/(2x-4), and m=-3/2, so we want to know if:

    d/dx[(x/2)+1/(2x-4)]=-3/2

    has any solutions.

    The detail of answering this have been dealt with by the other posters.

    RonL
    Last edited by CaptainBlack; October 4th 2006 at 08:53 PM.
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