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Math Help - proving local maximum and local minimum

  1. #1
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    proving local maximum and local minimum

    Suppose that f(x) and g(x) are positive functions such that f(x) has a local maximum and g(x) has a local minimum at x=c. Prove that

    h(x) = \frac{f(x)}{g(x)}

    this means if f'(c) = 0 and g'(c) = 0 that h'(c) = 0


    \frac{f'(x)g(x) - g'(x)f(x)}{[g'(x)]^2} at x = c \Rightarrow \frac{0g(x) - 0f(x)}{0^2} = \frac{0}{0} = 0

    or

    \frac{f'(c)}{g'(c)} = h'(c) \Rightarrow \frac{0}{0} = 0

    I don't know if I am close.
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  2. #2
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    Re: proving local maximum and local minimum

    or something along the lines if f(x) is divided by g(x) and is equal to h(x) when f(x) and g(x) have a local minimum and local maximum respectively that would make h(x) also have a local maximum at x = c because if h(x) = \frac{f(x)}{g(x} \Rightarrow g(x) = \frac{f(x)}{h(x)}
    Last edited by Jonroberts74; October 27th 2013 at 12:48 PM.
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    Re: proving local maximum and local minimum

    Quote Originally Posted by Jonroberts74 View Post
    Suppose that f(x) and g(x) are positive functions such that f(x) has a local maximum and g(x) has a local minimum at x=c. Prove that
    h(x) = \frac{f(x)}{g(x)}
    Therre is an open neighborhood, \mathcal{O} about c on which f(c) is an absolute maximum for f and on which g(c) is an absolute minimum for g.

    Thus, \forall x\in\mathcal{O} we know g(x)\ge g(c)>0 or \color{blue}0<\frac{1}{g(x)}\le\frac{1}{g(c)}.

    How can you use 0<f(x)\le f(c) with the above inequality?
    Last edited by Plato; October 27th 2013 at 01:52 PM.
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    Re: proving local maximum and local minimum

    That reply seems above my current skillset.
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    Re: proving local maximum and local minimum

    Quote Originally Posted by Jonroberts74 View Post
    That reply seems above my current skillset.
    OH! come on.
    Multiply \color{blue}0<\frac{1}{g(x)}\le\frac{1}{g(c)} through by f(x).
    Then replace with f(c) so that \frac{f(x)}{g(x)}\le \frac{f(c)}{g(c)} .
    Does that mean that h is bounded above on \mathcal{O}~?
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    Re: proving local maximum and local minimum

    Quote Originally Posted by Plato View Post
    OH! come on.
    Multiply \color{blue}0<\frac{1}{g(x)}\le\frac{1}{g(c)} through by f(x).
    Then replace with f(c) so that \frac{f(x)}{g(x)}\le \frac{f(c)}{g(c)} .
    Does that mean that h is bounded above on \mathcal{O}~?
    I don't know all the symbols in your reply. I will now take those replies and figure them out.

    I do not have a math professor teaching me nor have I had any classes. I do the best I can with the text books I have and use this forum to fill in the blanks.
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    Re: proving local maximum and local minimum

    Quote Originally Posted by Jonroberts74 View Post
    Suppose that f(x) and g(x) are positive functions such that f(x) has a local maximum and g(x) has a local minimum at x=c. Prove that

    h(x) = \frac{f(x)}{g(x)}

    this means if f'(c) = 0 and g'(c) = 0 that h'(c) = 0


    \frac{f'(x)g(x) - g'(x)f(x)}{[g'(x)]^2} at x = c \Rightarrow \frac{0g(x) - 0f(x)}{0^2} = \frac{0}{0} = 0

    or

    \frac{f'(c)}{g'(c)} = h'(c) \Rightarrow \frac{0}{0} = 0

    I don't know if I am close.
    I'm a bit confused here. What is h(x) supposed to be? Also you have two points where you say 0/0 = 0. 0/0 is undefined...it is not equal to 0.

    -Dan
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    Re: proving local maximum and local minimum

    Quote Originally Posted by Jonroberts74 View Post
    I do not have a math professor teaching me nor have I had any classes. I do the best I can with the text books I have and use this forum to fill in the blanks.
    Well that explains a lot
    Then you would not know that there are continuous functions that have no derivatives.
    Therefore, you cannot prove this using derivatives.

    To say that there is an open neighborhood, \mathcal{O}, of c means \mathcal{O}=(p,q) where p<c<q.
    In that open interval f(c) is an absolute maximum for f and g(c) is an absolute minimum for g.
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    Re: proving local maximum and local minimum

    Okay, I think I have some understanding. I understand the part of absolute maximum and absolute minimum. I still don't understand how this proves h(x) is equal to \frac{f(x)}{g(x)}.

    I will mark it down as a question I need to return to once I get a a stronger knowledge of the topic.


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