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Math Help - Advanced Calculus Question (analysis)

  1. #1
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    Advanced Calculus Question (analysis)

    f:R-->R and f '(x)<= a < 0 for any x

    show that f is onto R
    ---------------------------------------------------------------------------------------------------------------------
    Proof: since f '(x) < 0, we have f(x) a monotone decreasing function

    since f '(x) exists, we know f(x) must be continuous for any x in R

    the cardinality of two like sets is the same so: |R| = |R|

    Since the cardinality of the domain and range is the same, proving onto amounts to proving that f:R-->R is one-to-one

    But f(a)>f(b) for any a,b in R

    Thus, f:a-->f(a) and f:b-->f(b), and

    f(a) is not equal to f(b) whenever a is not equal to b

    which is to say f(x) is one-to-one for any x in R

    Since the cardinality of the domain equals the cardinality of the range (i.e. |R|=|R|) and f is one-to-one,

    f(x) must be ONTO.
    --------------------------------------------------------------------------------------------------

    Does anyone know what is wrong with this proof if anything? Thank you
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  2. #2
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    Re: Advanced Calculus Question (analysis)

    Quote Originally Posted by computerproof View Post
    f:R-->R and f '(x)<= a < 0 for any x

    show that f is onto R
    Consider $f(x)=-e^x-x$. Is that a counter-example?
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    Re: Advanced Calculus Question (analysis)

    Hi,
    The error in your "proof" is where you assert that if two sets A and B have the same cardinality, a function from A to B is onto if and only if f is one to one. Wrong, as Plato's example shows. The statement about onto and one to one is true only for finite sets A and B.
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    Re: Advanced Calculus Question (analysis)

    I don't see how it is a counterexample being that f '(x) is always negative forcing f(x) to be monotone decreasing. since f is defined for every possible value in the domain R, and f(x) is one-to-one: whenever a does not equal b we have f(a) does not equal f(b). With teh cardinality of the domain and range being the same I think this proves onto....but something still doesn't seem right...lol


    johng....i wrote this response before i had seen what you had written.....

    so the cardinality rule i state applies only to finite sets...ok

    can you explain the prposed counterexample....can't seem to wrap my mind around it
    Last edited by computerproof; February 26th 2014 at 10:01 AM.
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  5. #5
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    Re: Advanced Calculus Question (analysis)

    $\arctan(-x)$ is another

    it maps $(-\infty, \infty)$ to $\left(\frac{\pi}{2}, -\frac{\pi}{2}\right)$ yet is monotone decreasing over it's entire domain.

    So it meets your criterion of continuous and negative derivative but is not onto.
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    Re: Advanced Calculus Question (analysis)

    but arctan(-x) is bounded....since f '(x)<a<0, theis means f(x) < ax.....but since x can be any real number, f(x) can take on any value in R which is to say our f(x) is not bounded.....so i need to show f(x) is unbounded?
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  7. #7
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    Re: Advanced Calculus Question (analysis)

    can I use the fundamental theorem of calculs to say that this: f '(x) <=a<0

    implies this: f(x) <= ax

    (i basically integrated both sides
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  8. #8
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    Re: Advanced Calculus Question (analysis)

    you're right. I didn't catch the upper bound on f'
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    Re: Advanced Calculus Question (analysis)

    I feel unsure of myself when i use theorems because they do not always perfectly line up. for example the fundamental theorem of calculus is stated only for a finite interval....can i use it in cases like my question above where we are dealing with R, an infinite interval?
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  10. #10
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    Re: Advanced Calculus Question (analysis)

    I'm pretty sure that monotonicity and continuity imply the existence of an inverse of f.

    Given that proving f is onto is trivial. $\forall y \in Y\;\; f\left(f^{-1}(y)\right)=y$

    Isn't this easier than your current approach?
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  11. #11
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    Re: Advanced Calculus Question (analysis)

    Hi again,
    Your statement is definitely true. The attachment provides a proof.

    Advanced Calculus Question (analysis)-mhfcalc37.png
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