# Thread: proving local maximum and local minimum

1. ## 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.

2. ## 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 $h(x) = \frac{f(x)}{g(x} \Rightarrow g(x) = \frac{f(x)}{h(x)}$

3. ## Re: proving local maximum and local minimum

Originally Posted by Jonroberts74
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 with the above inequality?

4. ## Re: proving local maximum and local minimum

That reply seems above my current skillset.

5. ## Re: proving local maximum and local minimum

Originally Posted by Jonroberts74
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 $\mathcal{O}~?$

6. ## Re: proving local maximum and local minimum

Originally Posted by Plato
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.

7. ## Re: proving local maximum and local minimum

Originally Posted by Jonroberts74
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

8. ## Re: proving local maximum and local minimum

Originally Posted by Jonroberts74
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.
In that open interval $f(c)$ is an absolute maximum for $f$ and $g(c)$ is an absolute minimum for $g$.

9. ## 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.

Thank you