The Derivative

**zxcv** · March 21st 2009, 12:36 AM

May someone please show me how to proof the following:

Let f be differentiable on I. Show that if f is monotone increasing on I, then
f '(x) >= 0 for all x in I.

Note: >= is greater or equal to.

My idea:
Suppose that f is monotone increasing, that is f(x) <= f(c) for all x such that x<c. ... Then I know that after, we have to somehow say that [f(x)-f(c)]/(x-c) >= 0.
Can someone explain more and do we need any theorem to support the proof?
Thank you .

**CaptainBlack** · March 21st 2009, 12:50 AM

Originally Posted by zxcv

May someone please show me how to proof the following:

Let f be differentiable on I. Show that if f is monotone increasing on I, then
f '(x) >= 0 for all x in I.

Note: >= is greater or equal to.

My idea:
Suppose that f is monotone increasing, that is f(x) <= f(c) for all x such that x<c. ... Then I know that after, we have to somehow say that [f(x)-f(c)]/(x-c) >= 0.
Can someone explain more and do we need any theorem to support the proof?
Thank you .

Suppose that for some $x \in I\ f'(x)<0$ .

Let $h>0$ then as $f$ is monotone increasing:

$ \frac{f(x+h)-f(x-h)}{h} \ge 0 $

Hence:

$ \lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h} = f'(x) \ge 0 $

a contradiction, so there is no such point.

CB

**HeirToPendragon** · April 18th 2009, 09:06 PM

Originally Posted by CaptainBlack

Hence:

$ \lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h} = f'(x) \ge 0 $

a contradiction, so there is no such point.

Can you explain that a bit more. I'm getting lost in the letters.

**CaptainBlack** · April 18th 2009, 10:32 PM

Originally Posted by HeirToPendragon

Can you explain that a bit more. I'm getting lost in the letters.

If $f'(x)<0$ at some $x$ point in $I$ , then at that point (this uses the definition of having a derivative at that point):

$ \lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h} = f'(x) < 0 $

but we have just shown that as $h$ approaches $0$ from the left (using the assumption that $f$ is monotone increasing and the existance of the derivative at the point):

$ \lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h} = f'(x) \ge 0 $

which is a contradiction.

The only properties it uses are that $f$ is monotone increasing and the definition of a derivative.

CB

**HeirToPendragon** · April 19th 2009, 06:17 AM

I understand all that. What I'm confused on is how you know that f'(x) > 0

Is there a theorem out there or a general rule I'm not understanding?

We know that
$\frac{f(x+h)-f(x-h)}{h} \ge 0$

But how does that prove that
$\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h} = f'(x) \ge 0$
What is the reason that f'(x) can't be < 0 there?

**CaptainBlack** · April 19th 2009, 08:01 AM

Originally Posted by HeirToPendragon

I understand all that. What I'm confused on is how you know that f'(x) > 0

Is there a theorem out there or a general rule I'm not understanding?

We know that
$\frac{f(x+h)-f(x-h)}{h} \ge 0$

But how does that prove that
$\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{h} = f'(x) \ge 0$
What is the reason that f'(x) can't be < 0 there?[/font]

If $f(x)$ is monotone increasing what is the sign of:

$\frac{f(x+h)-f(x-h)}{h}$

when $h>0$ ?

CB

**HeirToPendragon** · April 19th 2009, 08:09 AM

What do you mean the sign?

Do you mean is it positive or negative?

It's positive. That makes perfect sense. But what I don't understand is why you know the lim of it is also positive.

If a(x) is positive does that mean that it's lim has to be?

**CaptainBlack** · April 19th 2009, 08:22 AM

Originally Posted by HeirToPendragon

What do you mean the sign?

Do you mean is it positive or negative?

It's positive. That makes perfect sense. But what I don't understand is why you know the lim of it is also positive.

If a(x) is positive does that mean that it's lim has to be?

No it means the limit cannot be negative.

CB

**HeirToPendragon** · April 19th 2009, 08:25 AM

Ok so what is the proof/reasoning for that?

I don't mean to sound annoying, but I really want to know why I'm allowed to just say this.

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