derivative inequality

BrainMan

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Joined
Oct 24, 2007
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26
I'm trying to think this out, but I just can't get it to make any sense:

Let f be continuous on [a,b] (a<b) and differentiable on (a,b). If f(a) = 0 and abs(f'(x)) <= c*abs(f(x)) on (a,b), show that abs(f'(x)) <= (c^2)*(x-a)*abs(f(d)), a < d < x.

Does anyone know what is needed to be done here? Thanks in advance.
 
I think I can use the Mean Value Theorem in the form f(x) = f'(d)(x-a) where a < d < x. OK, so I get:
abs(f'(x)) <= c*abs(f'(d)(x-a)) = c*(x-a)*abs(f'(d)). What do I do now?
 
Forget what I wrote before. I just need to:

prove that f = 0 on [a,b] given the conditions at the beginning, namely:
-f continuous on [a,b] (a<b)
-f differentiable on (a,b)
-f(a) = 0
-abs(f'(x)) <= c*abs(f(x)) on (a,b)

How would I do this? Thanks in advance.
 
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