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Math Help - Roots of a function

  1. #1
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    Roots of a function

    I help some help please.

    Basically if i had a graph of e.g.

    y=-(x-3)^2+1

    I know that my graph would intersect the x-axis at x=2 and x=4.

    However if I didn't know the equation of my curve, but knew the shape of the graph was like y=-(x-3)^2+1, and it crossed the x-axis at 2 points.

    How could I prove that it had 2 roots?

    I think I may have something to do with continuity, but not sure.

    Any help is appreciated.
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  2. #2
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    I'm not sure I understand the problem. Crossing the x-axis at 2 points means the same thing as having 2 roots, doesn't it?

    If you have a continuous function f(x), and it is positive for x=x_0, negative for x=x_1, and then positive again for x=x_2\text{ (with }x_0<x_1<x_2\text{)}, then you know that it has at least 2 roots.

    If you can tell me more, I might be able to help more. If the problem is from a book or problem set, it usually helps to copy the whole problem word-for-word. It also helps to know what you're studying.

    Post again in this thread and I'll answer as soon as I can.

    - Hollywood
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  3. #3
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    Basically I have the function v(u) = -u + 1/RoInU + C where C is a constant of integration.

    I need to determine relevant solutions so that v(u)=0

    I already know that v(1)=0

    I need to sketch the curves with 0.1<=u<=1 for an illustrative range of values of Ro.

    Out of these curves could also be traced out during an epidemic.
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  4. #4
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    So you have a function v(u)=-u+\frac{1}{R_0I_nu}+C and you need to find where v(u)=0.

    You can use the boundary condition v(1)=0 to determine C:

    v(1)=-1+\frac{1}{R_0I_n}+C=0\text{, so }C=1-\frac{1}{R_0I_n}

    Substituting back in to the equation gives:

    v(u)=-u+1+\frac{1}{R_0I_n}\left(\frac{1}{u}-1\right)

    And the roots are given by:

    v(u)=-u+1+\frac{1}{R_0I_n}\left(\frac{1}{u}-1\right)=0

    1-u+\frac{1}{R_0I_n}\left(\frac{1-u}{u}\right)=0

    We already know about the root at u=1, so we can cancel 1-u, giving:

    1+\frac{1}{R_0I_n}\left(\frac{1}{u}\right)=0

    u=-\frac{1}{R_0I_n} as the second root.

    I don't know if the combination R_0I_n can be negative. If it has to be positive, and assuming u is always positive, then there is only the one root at u=1. If it can be negative, there is a second root.

    I hope that was helpful. Post again in this thread if you're still having trouble.

    - Hollywood
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