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Thread: Intermediate Value Theorem

  1. #1
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    Intermediate Value Theorem

    Let f: [0, +oo] of all real numbers be differentiable with f(0)=A>0 and f'(x)<= 0 for all x>0

    (a) show that there exists some c of (0, +oo) such that f(c)=c
    (b) show that there is only one such c

    (Note: These facts are obviously intuitively, as you can see be drawing a graph showing the curve y=f(x) and the 45-degree line y=x. The problem is to prove them analytically, using theorems. Since f'<= 0, what do we know about f? Let g(x)=f(x)-x. What do we know about g(0) and g(2A)? What do we know about g'? )
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  2. #2
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    Re: Intermediate Value Theorem

    Quote Originally Posted by Headinthesand View Post
    Let f: [0, +oo] of all real numbers be differentiable with f(0)=A>0 and f'(x)<= 0 for all x>0

    (a) show that there exists some c of (0, +oo) such that f(c)=c
    (b) show that there is only one such c
    Note that one never writes $[a,\infty\large]$ Because infinity is not a number we must write it as $[a,\infty\large)$.

    Now consider the problem.
    Example: $\large f(x)=e^{-x}$ now note that:
    1) $f$ is differentiable on $[0,\infty)~?$
    2) $f(0)=1>0~?$
    3) is it true that $\left( {\forall x} \right)\left[ {f'(x) < 0} \right]?$
    4 look at the graph; is the graph ever negative?

    What can we say about the question?
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  3. #3
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    Re: Intermediate Value Theorem

    The way you stated the problem, I'm not sure if the domain of f is all of the reals. In any case, for your conclusions to be true, you need (right) continuity of f at 0. Easy counterexamples for an f with domain $[0,\infty)$.

    You were on the right track to consider $g(x)=f(x)-x$. Since $g$ is continuous and $g(0)>0$, all you need is some $x>0$ with $g(x)\leq0$. Suppose $g(A)>0$, but this contradicts the fact that $f$ is non-increasing.

    Next, to prove the uniqueness of c, just prove that $g$ is strictly decreasing.
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