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Math Help - proof: working with derivatives, possibly mean value theorem?

  1. #1
    Senior Member sfspitfire23's Avatar
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    proof: working with derivatives, possibly mean value theorem?

    Let f:R-->R be differentiable for all x. Suppose that f'(x)>1 for all x. Show that the graph of y=f(x) can intersect the graph of y=x no more than once.

    I have no idea where to even begin... any ideas?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Let f:R-->R be differentiable for all x. Suppose that f'(x)>1 for all x. Show that the graph of y=f(x) can intersect the graph of y=x no more than once.

    I have no idea where to even begin... any ideas?
    First you need to realize what the question is asking.

    If \Gamma_f (the graph) intersects \Delta=\left\{(x,x):x\in\mathbb{R}\right\} then that means that (x,f(x))=(x,x)\implies f(x)=x. So, why can't we have that f(x)=x,f(x')=x' for x\ne x'?
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Isn't x' just going to be zero, though, seeing as x is any real number?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Isn't x' just going to be zero, though, seeing as x is any real number?
    I have no idea what you mean. This is much, much simpler than you're making it. The question is asking why a function whose derivative which is strictly greater than one can't have two fixed points. To see this suppose that it did, call them x_1,x_2,\text{ }x_1<x_2. Then, what does the MVT say about the derivative of f on [x_1,x_2]?
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  5. #5
    Senior Member sfspitfire23's Avatar
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    Oh I see now. So if we suppose towards contradiction that f has two fixed points: x_1 and x_2 with x_1< x_2, then we would have f(x_1)= x_1 and f(x_2)= x_2. Then, by the Mean Value Theorem, we would have for some c in the interval (x_1, x_2):
    f'(c) = f(x_2) - f(x_1) / x_2 - x_1 = (x_2 - x_1)/(x_2 - x_1) = 0
    which contradicts the assumption that f'(x)>1 for all x...?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Oh I see now. So if we suppose towards contradiction that f has two fixed points: x_1 and x_2 with x_1< x_2, then we would have f(x_1)= x_1 and f(x_2)= x_2. Then, by the Mean Value Theorem, we would have for some c in the interval (x_1, x_2):
    f'(c) = f(x_2) - f(x_1) / x_2 - x_1 = (x_2 - x_1)/(x_2 - x_1) = \color{red}1
    which contradicts the assumption that f'(x)>1 for all x...?
    Yes, except I'm sure you meant what I changed above
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  7. #7
    Member mohammadfawaz's Avatar
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    In more general terms, given f'(x)\neq 1, you can have at most one fixed point. But if you also have |f'(x)|\leq A<1, then you know for sure that there exists a fixed points, and hence it is unique. To show this, take the sequence defined by x_{n+1}=f(x_n) and show it is convergent to the unique fixed points.
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