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Math Help - Show that the tangent lines of two different functions are parallel

  1. #1
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    Show that the tangent lines of two different functions are parallel

    I have this problem which I don't know how to even start. Can anyone help me solve this problem?



    Would really appreciate any help. Thanks!
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  2. #2
    Super Member TheChaz's Avatar
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    Quote Originally Posted by colerelm1 View Post
    I have this problem which I don't know how to even start. Can anyone help me solve this problem?



    Would really appreciate any help. Thanks!
    1. What function can we use to represent the vertical distance?
    2. How do you find the greatest value of a function (maximum)? Remember; this is calculus...
    3. What would it mean for the tangents to be parallel?

    If you can sufficiently answer these questions, you could submit that as your solution, with little supplementary justification.
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    Ok so I know that f'(x) = 0 at c and g'(x) = 0 at c. I also know that since these derivatives are equal, the slope of their respective functions must be equal at c.

    I'm confused as to how I'm supposed to make the assumption that, because the distance is the greatest between the two functions at c, then the derivatives must be equal, making the slope of the tangent curves parallel.
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by colerelm1 View Post
    Ok so I know that f'(x) = 0 at c and g'(x) = 0 at c. I also know that since these derivatives are equal, the slope of their respective functions must be equal at c.
    ...
    You do not know that! Those curves don't (both) have horizontal slopes anywhere!! Furthermore, f'(c) = 0 = g'(c) (which is another way of saying what you said) is what we are trying to prove - at least the f'(c) = g'(c) part.

    I'll answer my own questions.

    1. Let's call the distance between the curves, as a function of x, D(x).
    D(x) = g(x) - f(x).
    2. (No.... you do this part)
    3. Then we get exactly what we want --- g'(c) = f'(c). Matter 'o fact, you could just subtract f'(c) from both sides and get the missing step...
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    oops...I don't know why I said that the derivatives at c are equal to zero....I know they're not. I meant g'(x)=f'(x). I'm going to look it over again, and with your help, I think I have it almost completed. Thanks a lot bro.
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  6. #6
    Super Member TheChaz's Avatar
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    Quote Originally Posted by colerelm1 View Post
    oops...I don't know why I said that the derivatives at c are equal to zero....I know they're not. I meant g'(x)=f'(x). I'm going to look it over again, and with your help, I think I have it almost completed. Thanks a lot bro.
    I'm telling you; the ENTIRE solution is right under your nose! Here's what I would put:

    "Let D(x) = g(x) - f(x) represenent the vertical distance between g and f. Since D is maximized at c, we know that ___________. This implies that g'(c) = f'(c) <=> the tangents to the curves are parallel, as desired"
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    Quote Originally Posted by TheChaz View Post
    I'm telling you; the ENTIRE solution is right under your nose! Here's what I would put:

    "Let D(x) = g(x) - f(x) represenent the vertical distance between g and f. Since D is maximized at c, we know that ___________. This implies that g'(c) = f'(c) <=> the tangents to the curves are parallel, as desired"
    we know that G is minimized at C?
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    Ahem!

    Take a moment to think about what Chaz has been trying to get you to realize! First, there is no function G. If you mean g(x) is minimized, then clearly you'd be wrong; take a look at the graph and notice that there is no point where either function attains a maximum or minimum on the open interval (a,b).

    Consider, as Chaz suggested you do, the function D(x) = g(x) - f(x) which gives the vertical distance between two auxiliary functions g(x) and f(x) depicted in the graph. The problem states that D(x) is maximized at (c, D(c)) -- e.g., D(c) = g(c) - f(c) = maximum value that D obtains. Note that this condition does not necessarily mean that f(x) or g(x) are maximized or minimized (or vice versa) at 'c', nor does it matter for this problem.

    Now, if D(x) = g(x) - f(x) is maximized at c, then consider what is then implied about D'(x) evaluated at 'c'. While you consider this, keep in mind the linearity property of differentiation, and you should see immediately that f'(c) = g'(c). From this equivalence, deduce that f(x) and g(x) have parallel (not necessarily equal) tangent lines at point c.

    It may be especially helpful for you to reconsider the geometric meanings of functions and their first and second derivatives, in addition to what geometric conditions must be satisfied for two lines to be (a) parallel and (b) perpendicular (and how these conditions relate to the derivatives of respective functions).
    Last edited by TaylorM0192; April 19th 2011 at 09:27 PM.
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  9. #9
    Super Member TheChaz's Avatar
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    Quote Originally Posted by colerelm1 View Post
    we know that G is minimized at C?
    I'll just spell it out for the sake of completion.

    D'(c) = 0 means - BY DEFINITION!
    g'(c) - f'(c) = 0

    Add f'(c) to both sides and you're done.
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