# 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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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

Staff Emeritus
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.