# Thread: mean value theorem section problem

1. ## mean value theorem section problem

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 $\frac{f(x)}{x}\leq max{|f(x)|:0<=x<=a}$
(iii)f(x)=0 for $x\in[0,a]$
(iiiן)f(x)=0 for $x\in[0,1]$
in each of the following subquestion we can use the previosly proves subquestion.

first part i have solve by saying that f is continues in the subsection so
by weirshtrass we have max and min
and max|f(x)|=max{|maxf(x)|,|minf(x)|}

in the second part
we know that max|f(x)|>|f(x)|>=|f'(x)|
and we take c in [0,x] a subsection of [0,a]
|f(c)|>=|f'(c)|
and we know that f(0)=0 so we take [0,a]
|f'(c)|=|f(0)-f(x) /x-0 |
|f'(c)|=|f(x)/x|
|f(x)|>|f(x)|>=|f'(x)|
so i got all the parts but i cant join them because its c there and not x
c is inside point x is on the border.

what to do?

2. ## Re: mean value theorem section problem

By MVT, $\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=f'(c)$ for some $c\in(0,x)$. Now, $f'(c)\le|f'(c)|\le |f(c)|$ by assumption, and $|f(c)|\le\max\{|f(x)|:0\le x\le a\}$ since $x\in(0,a]$.

Hint: use \{ and \} to make curly braces show in LaTeX.