f is continues in [0,1] and differentiable in (0,1)(adsbygoogle = window.adsbygoogle || []).push({});

f(0)=0 and for[TEX] x\in(0,1)[/TEX] [TEX]|f'(x)|<=|f(x)|[/TEX] and 0<a<1 prove:

(i)the set [TEX]{|f(x)| : 0<=x<=a}[/TEX] has maximum

(ii)for every [TEX]x\in(0,a][/TEX] 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]

(iiiן)f(x)=0 for [TEX]x\in[0,1][/TEX]

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?

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# Homework Help: Mean value theorem section problem

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