# Thread: A couple of proofs using Mean Value Theorem

1. ## A couple of proofs using Mean Value Theorem

I've gotten through the entirety of this assignment and about 6 other proofs without any problem, but I'm just struggling with these two. Both of these are in a section that heavily covers the Mean Value Theorem. I'm not looking for answers directly, but maybe pointers could be nice. All of my friends are artsy and avoid math like the plague.

(Edit here.. I had the problem written down slightly wrong.)

If f is a differentiable, odd function show that ∀b>0, ∃c∈(-b,b) st f'(c) = f(b)/b
I've gotten as far as..
f(-b) - f(b) = f'(c)(-b - b)
f'(c) == (f(-b) - f(b))/(-b - b)

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I've gotten the second. So others may take a look at this later for any reason, I'll leave my work here:

Prove |sin a - sin b| ≤ |a - b|

sin a - sin b = d/dc(sin c)(a - b) by Mean Value Theorem
sin a - sin b = cos c (a - b)
|sin a - sin b| = |cos c||a - b|
|sin a - sin b| ≤ |a - b| since |cos c| >= 0

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If I need to post these both on different threads, I'll do so. I was a bit unclear about the section in the rules saying new questions need new threads, and if that means that two questions had to have their own.

Thank you

2. Originally Posted by Nikker
I've gotten through the entirety of this assignment and about 6 other proofs without any problem, but I'm just struggling with these two. Both of these are in a section that heavily covers the Mean Value Theorem. I'm not looking for answers directly, but maybe pointers could be nice. All of my friends are artsy and avoid math like the plague.

I've gotten as far as..
f(-b) - f(b) = f'(c)(-b - b)
f'[c] == (f(-b) - f(b))/(-b - b)

I'm assuming I have to somehow use the the fact that f is an odd function to show that (f(-b) - f(b))/(-b - b) is equivalent to f'(b)/b... But I'm not sure what steps are in between.
Are you sure that the problem states $\displaystyle f'(c)=\frac{f'(b)}{b}$? What if $\displaystyle f(x)=x$ and $\displaystyle b=\frac{1}{2}$?

The second is to prove that |sin a - sin b| ≤ |a - b|

And all I have is this:
sin a - sin b = d/dc(sin c)(a - b)
sin a - sin b = cos c (a - b)
For the next step, note that

$\displaystyle |\sin a - \sin b|=|\cos c||a-b|.$

3. Ahhh, you're right. I'm not sure where that came from, but it's f(b)/b, not f'(b).

Thanks. I'll update the question and take a closer look.

Edit: Your tip also definitely helped with the second one.

I'm still not much further with the first, since I know I have to use the fact that it's an odd function, but am not quite sure where.

4. $\displaystyle \frac{f(-b)-f(b)}{-b-b}=\frac{-f(b)-f(b)}{-b-b}=\frac{f(b)}{b}.$

5. Originally Posted by Krizalid
$\displaystyle \frac{f(-b)-f(b)}{-b-b}=\frac{-f(b)-f(b)}{-b-b}=\frac{f(b)}{b}.$
And we know that f(-b) = -f(b) because f is an odd function?

6. yes.

7. Awesome. Thank you very much.

It's always the small things I miss, apparently.