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?
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 am going to restate a problem a little bit because I usually use the same notation over and over again.
1)Let be a continous function on differenciable on .
2)Let .
3)There exists so that for all .
To show: For any we can choose so that .
Proof: By #1 use the mean value theorem on on the interval that means for some . Using #2 we can write thus . By #3 we know by what we just got it means . Q.E.D.
I also have to show that
abs(f(x)) <= (c^n)*[(x-a)^n]*M where M = max(abs(f(x))) on [a,b].
I'm going to use my notation:
I use the MVT f(x) = f'(d)(x-a).
Then, abs(f(x)) <= abs(f'(d))*(x-a) <= c*abs(f(d))*(x-a) <= c(x-a). I understand that M will be greater than or equal to all abs(f(x)), but how do I get the c and the (x-a) to the nth power?