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Math Help - analysis question

  1. #1
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    analysis question

    Hi. I'm taking an Analysis I course and we just started differentiation. I'm not 100% comfortable with it yet and would appreciate some help on a problem in my textbook. Suppose that f:[0,1] to R is continuous on f(0)= 0, differentiable for x in [0,1], and 0 <= f '(x) <= af(x) for a > 0. Prove that f = 0. (Hint: consider the derivative of (e^(-ax))f(x).) Any help you offer would be greatly appreciated.
    Last edited by agentZERO; November 17th 2007 at 12:50 PM.
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  2. #2
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    Please reread what you have posted.
    Is it exactly as it appears in your text book? (What is the text?)
    It just does not seem to be correct.
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  3. #3
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    It has that f --- 0 where --- is three horizontal lines on top of each other (basically an equal sign with an extra line). I figured this meant equals, but I guess it does not. What exactly does it mean and how would you solve this? Thank you for trying to help.
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  4. #4
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    Quote Originally Posted by agentZERO View Post
    It has that f --- 0 where --- is three horizontal lines on top of each other (basically an equal sign with an extra line). I figured this meant equals
    That is correct, but that is not what I asked.
    What does "f:[0,1] to R is continuous on f(0)= 0" mean?
    I suppect that it is not written that way in the text.
    What is the textbook you are using?
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  5. #5
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    Yikes. I did make a serious mistake. It should be: Suppose that f:[0,1] to R is continuous with f(0)= 0, differentiable for x in [0,1], and 0 <= f '(x) <= af(x) for a > 0. Prove that f = 0. (Hint: consider the derivative of (e^(-ax))f(x).)
    I'm using a textbook outside of class just to study. It's called Elementary Real Analysis by Brian S. Thomson.
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  6. #6
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    Use the hint:
     - ae^{ - ax} f(x) + e^{ - ax} f'\left( x \right) = e^{ - ax} \left( { - af(x) + f'\left( x \right)} \right) \le e^{ - ax} \left( { - af(x) + af\left( x \right)} \right)<br />
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  7. #7
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    Ah, that makes sense. How do you know f is zero though?
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  8. #8
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    Do you want to discover anything about this problem for yourself?
    Does that mean that the derivative of e^{ - ax} f(x) is zero everywhere?
    What does that mean, recalling f(0)=0?
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  9. #9
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    You don't have to be a grouch. You're the one helping and I thought you wanted to help. Plus, it's not always easy for people who don't know the code to type out their work. It doesn't mean they're not thinking and trying to work it out themselves.
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  10. #10
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    Quote Originally Posted by agentZERO View Post
    You don't have to be a grouch. You're the one helping and I thought you wanted to help. Plus, it's not always easy for people who don't know the code to type out their work. It doesn't mean they're not thinking and trying to work it out themselves.
    He isn't being a grouch. He simply wants to see evidence that you are doing something to try to solve the problem. Hence even if you have no work to show, you could mention what you have tried to do to solve the problem.

    And, as most members here don't know how to use LaTeX, you can post your work in the following kind of format:
    -ae^{ - ax} f(x) + e^{ - ax} f'(x) = e^{ - ax} (- af(x) + f'(x)} ) <= e^{ - ax} ( - af(x) + af(x))

    It's harder to read, but still quite understandable.

    -Dan
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