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Math Help - Help! The middle value theorem

  1. #1
    Senior Member Shanks's Avatar
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    Help! The middle value theorem

    f(x) is continous in [0,1], differentiable in (0,1), and f(0)=0,f(1)=1.
    prove that for any positive numbers a and b, there exist < in (0,1) such that
    .

    P.S I have posted this thread in the calculus subforum, but it is unsolved yet. I just want somebody help me out.
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  2. #2
    Senior Member Shanks's Avatar
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    We have never met problem like this, (usually we are supposed to provethat there is a point in the open interval satisfies certain condition which contains only one parameter x ).
    It may be a chanllenge problem, please help me out.
    Thank you more than I can say!
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  3. #3
    MHF Contributor Drexel28's Avatar
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    What is the middle value theorem? That may give some insight as to how to solve this. The x_1<x_2 is what makes this bothersome of course, for if that stipulation wasn't in place we know there exists some c\in(0,1) s.t. f'(c)=\frac{1-0}{1-0}=0. Which of course taking x_1=x_2=c makes this trivial. I'll give it more thought.
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  4. #4
    Senior Member Shanks's Avatar
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    Quote Originally Posted by Drexel28 View Post
    What is the middle value theorem? That may give some insight as to how to solve this. The x_1<x_2 is what makes this bothersome of course, for if that stipulation wasn't in place we know there exists some c\in(0,1) s.t.
    f'(c)=\frac{1-0}{1-0}=0" alt="
    f'(c)=\frac{1-0}{1-0}=0" />. here 1, not 0.

    Which of course taking x_1=x_2=c makes this trivial. I'll give it more thought.
    A type error!
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Shanks View Post
    A type error!
    Ok? Sweet. That was an obvious typo. That does not in anyway answer the question.
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  6. #6
    Senior Member Shanks's Avatar
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    I mean the intermediate value theorm.
    we can give a little manipulation to the representation, that is equivalent to show that
    for any positive number \lambda in (0,1), there exist x_1<x_2,\text{ such that } <br />
\frac{\lambda}{f'(x_1)}+\frac{1-\lambda}{f'(x_2)}=1
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  7. #7
    Senior Member Shanks's Avatar
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    Is this statement true or false?
    "suppose f is continous in [a,b], differentiable in (a,b). If r and s in (a,b) satisfy f'(r)<f'(s), then for any point c in the open interval ( f'(r), f'(s) ), there exist a point t in (a,b) such that f'(t)=c."
    I think it is true! and we may apply it to the prove. what is your opinion?
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  8. #8
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    Quote Originally Posted by Shanks View Post
    f(x) is continous in [0,1], differentiable in (0,1), and f(0)=0,f(1)=1.
    prove that for any positive numbers a and b, there exist < in (0,1) such that
    .
    By the mean value theorem, there is a point c with 0<c<1 such that f'(c)=1. If there is more than one such point, say c<d and f'(d)=1 then clearly we can take x_1=c and x_2=d. So we may assume that there is a unique point c with f'(c)=1.

    It cannot happen that f(c)=c (because then the mean value theorem applied to the interval [0,c] would lead to another point where the derivative is 1. So suppose that c<f(c)<1. (A similar proof would deal with the case 0<f(c)<c.)

    The idea now is to draw a line through the point (c,f(c)) with gradient less than 1 but large enough that its value at x=1 is larger than 1. This line must cross the graph of f(x) at some point between c and 1. To do that, choose \delta>0 so that \delta<\frac{f(c)-c}{a(1-f(c))}. Then the line y-f(c) = \frac{x-c}{1+a\delta} will have those properties. In fact, if we define g(x) = f(x) - f(c) - \frac{x-c}{1+a\delta} then you can verify that g'(c)>0 and g(1)<0. So g(x) must be positive when x is slightly larger than c, and g(x) is negative when x=1. By the intermediate value theorem there is a point q with c<q<1 such that g(q)=0. Then by Rolle's theorem there is a point x_2 with c<x_2<q with g'(x_2)=0. Therefore f'(x_2) = \frac1{1+a\delta}.

    Now we do something similar in the interval [0,c]. This time, we want a line through (c,f(c)) with gradient greater than 1 that crosses the y-axis above the origin. To do that, we choose \delta<\frac{f(c)-c}b. Then the line y-f(c) = \frac{x-c}{1-b\delta} has the required properties, and it must cross the graph of f(x) at some point p between 0 and c. A similar argument to the previous paragraph shows that there is a point x_1 with 0<x_1<c such that f'(x_1) = \frac1{1-b\delta}.

    Then \frac a{f'(x_1)} + \frac b{f'(x_2)} = a(1-b\delta) + b(1+a\delta) = a+b.

    [If we were told that f'(x) is continuous then the above argument could be made a lot shorter, by applying the intermediate value theorem to f'(x). But we are not told that, and so the only tool available is to use the mean value theorem on f(x).]
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  9. #9
    Senior Member Shanks's Avatar
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    I think your prove have some flaw.
    what about f(c)>1?
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  10. #10
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    Quote Originally Posted by Shanks View Post
    I think your prove have some flaw.
    what about f(c)>1?
    I had overlooked that possibility. But in fact the proof becomes easier in that situation, because \delta can then take any positive value and the rest of the argument works as before.

    I was not trying to provide a watertight analytic proof of this rather tricky result, but to give an indication of the geometric reason why it is true. If f(c)>1 then any line through (c,f(c)) with positive gradient less than 1 will cross the graph of f(x) between x=c and x=1.
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  11. #11
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    Quote Originally Posted by Opalg View Post
    If we were told that f'(x) is continuous then the above argument could be made a lot shorter, by applying the intermediate value theorem to f'(x). But we are not told that, and so the only tool available is to use the mean value theorem on f(x).
    Actually you don't need that f'(x) is continuous. It is know that if f'(a)<\lambda<f'(b) then there exist a c\in(a,b) such that f'(c)=\lambda
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  12. #12
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    Quote Originally Posted by putnam120 View Post
    Actually you don't need that f'(x) is continuous. It is know that if f'(a)<\lambda<f'(b) then there exist a c\in(a,b) such that f'(c)=\lambda
    Thanks, that's something I should have known (probably did, at some time in the remote past, but my memory's not what it was). In fact, I think that the technique I used in the above comment would probably also prove that interpolation property for the derivative.
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  13. #13
    Senior Member Shanks's Avatar
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    So...the statement is true. right?
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