Homework Help: Differentiability and extreme points question

1. Sep 6, 2011

nhrock3

2.b)
f is continues in [0,1] and differentiable in (0,1)
f(0)=0 and for $$x\in(0,1)$$ |f'(x)|<=|f(x)| and 0<a<1
prove:
(i)the set {|f(x)| : 0<=x<=a} has maximum
(ii)for every x\in(0,a] this innequality holds [TEX]\frac{f(x)}{x}\leq max{|f(x)|:0<=x<=a}[/TEX]
(iii)f(x)=0 for [TEX]x\in[0,a][/TEX]
(iii)f(x)=0 for [TEX]x\in[0,1][/TEX]
in each of the following subquestion we can use the previosly proves subquestion.

2. Sep 6, 2011

micromass

So what did you try?? If you tell us what you tried, then we'll know where to help?

3. Sep 6, 2011

nhrock3

i dont want solution only starting guidence.
for 1 i know the a continues function has a maximum

4. Sep 6, 2011

micromass

If you want guidance, then you'll have to show us what you've tried. You won't get a solution here...

5. Sep 6, 2011

nhrock3

there is a difference between solution
and starting guidence

6. Sep 6, 2011

micromass

For 1, you indeed use that a continuous function has a maximum. What continuous function do they think they're using here??

7. Sep 6, 2011

nhrock3

i only know about the function what was given

8. Sep 6, 2011

micromass

What function do you need to show continuity of??

9. Sep 6, 2011

nhrock3

f is continues in [0,1] and differentiable in (0,1)
f(0)=0 and for $$x\in(0,1)$$ |f'(x)|<=|f(x)| and 0<a<1

10. Sep 6, 2011

micromass

Yes, that is what we are given.

But we need to prove in (1) that

$$\{|f(x)|~\vert~x\in [0,a]\}$$

has a maximum. To show this, we must use that every continuous function on a closed interval has a maximum.
So, what will we take as our continuous function?

11. Sep 6, 2011

nhrock3

we cant take a spesific function
its a proof for all function

a proof for a spesific is not sufficient

12. Sep 6, 2011

Tomer

You don't understand what micromass is trying to imply. He's not suggesting taking a specific function like f(x) = sinx. He means you need to choose the correct function in your proof, which has to do with f(x) (hint hint), in order to prove that {|f(x)| | 0<= x <= a} has a maximum.
The answer is pretty clear - you just need to formulate a short proof.
Get it?

13. Sep 6, 2011

nhrock3

ahh now i get it
we need to formulate some other functionand using it we prove about
our abstract f(x)

maybe g(X)=|f(x)|
?

14. Sep 6, 2011

micromass

Yes, so prove that |f(x)| is continuous if f is.

15. Sep 6, 2011

nhrock3

ionly know one way
and it showing that $$lim_{x->x0}f(x)=f(x0)$$

but its not possible
because we dont have an actual function

16. Sep 6, 2011

micromass

Can you show |x| to be continuous??

Using that, you can show |f(x)| to be continuous as composition of f(x) and |x|.

17. Sep 7, 2011

Tomer

You don't need to show that, it's a given.
Like said above me, asking if the set {|f(x)| | a<=x<=b} has a maximum is just like asking "does the function |f(x)| (notice the absolute value!) get a maximum in the segment [a,b]?". You know there would be an easy answer if you'd say "|f(x)| is continuous on this closed segment and therefore gets a maximum". Therefore, you need to show that |f(x)| is a continuous.