1. ## The Derivative

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 .

2. 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

3. 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.

4. 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
$

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

CB

5. 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?

6. 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

7. 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?

8. 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

9. 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.