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Math Help - Using IVT (Int Value Therom) on a Hill

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    Using IVT (Int Value Therom) on a Hill

    Using IVT (Int Value Therom) on a Hill-calculus-continuous-problem.png
    (Click the picture to see the problem and what I have done so far)
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    Re: Using IVT (Int Value Therom) on a Hill

    Quote Originally Posted by Chaim View Post
    Click image for larger version. 

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    (Click the picture to see the problem and what I have done so far)
    So you have defined two different functions. You have

    s:[0,20] \to [0,1]

    and

    r:[0,10] \to [0,1]

    For these functions to be compatable they must agree on where you define the top and the bottom.

    If you defince the top to 1 and the bottom to be 0 then you must have these values

    s(0)=0 \quad s(20)=1

    and

    r(0)=1 \quad r(10)=0

    So you get the two functions

    s(t)=\frac{t}{20} \quad r(t)=1-\frac{t}{10}

    Now if you define their difference it will only exist on the intersection of their domains!

    f:[0,10] \to [-1,.5] \quad f(t) =s(t)-r(t)=\frac{3t}{20}-1

    But notice that

    f(0) =-1 \text{ and } f(10)=.5

    Since f is continous the intermediate value theorem says there exists a c \in [0,10] such that f(c)=0

    So we get that

    f(c)=s(c)-r(c) =0 \iff 0=s(c)-r(c) \iff s(c)=r(c)

    So the hiker was at the same place at the same time.

    Note: Since the function is linear would could solve for the value of t explicitly, but the power of the IVT is we don't have to be able to find it.
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