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Thread: 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

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

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


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

    or

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

    I don't know if I am close.
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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 $\displaystyle h(x) = \frac{f(x)}{g(x} \Rightarrow g(x) = \frac{f(x)}{h(x)}$
    Last edited by Jonroberts74; Oct 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
    $\displaystyle h(x) = \frac{f(x)}{g(x)}$
    Therre is an open neighborhood, $\displaystyle \mathcal{O}$ about $\displaystyle c$ on which $\displaystyle f(c)$ is an absolute maximum for $\displaystyle f$ and on which $\displaystyle g(c)$ is an absolute minimum for $\displaystyle g$.

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

    How can you use $\displaystyle 0<f(x)\le f(c)$ with the above inequality?
    Last edited by Plato; Oct 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 $\displaystyle \color{blue}0<\frac{1}{g(x)}\le\frac{1}{g(c)}$ through by $\displaystyle f(x)$.
    Then replace with $\displaystyle f(c)$ so that $\displaystyle \frac{f(x)}{g(x)}\le \frac{f(c)}{g(c)}$ .
    Does that mean that $\displaystyle h$ is bounded above on $\displaystyle \mathcal{O}~?$
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    Re: proving local maximum and local minimum

    Quote Originally Posted by Plato View Post
    OH! come on.
    Multiply $\displaystyle \color{blue}0<\frac{1}{g(x)}\le\frac{1}{g(c)}$ through by $\displaystyle f(x)$.
    Then replace with $\displaystyle f(c)$ so that $\displaystyle \frac{f(x)}{g(x)}\le \frac{f(c)}{g(c)}$ .
    Does that mean that $\displaystyle h$ is bounded above on $\displaystyle \mathcal{O}~?$
    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

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

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


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

    or

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

    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, $\displaystyle \mathcal{O}$, of $\displaystyle c$ means $\displaystyle \mathcal{O}=(p,q)$ where $\displaystyle p<c<q$.
    In that open interval $\displaystyle f(c)$ is an absolute maximum for $\displaystyle f$ and $\displaystyle g(c)$ is an absolute minimum for $\displaystyle g$.
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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 $\displaystyle h(x) $ is equal to $\displaystyle \frac{f(x)}{g(x)}$.

    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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