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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Mean value theorem section problem

**Physics Forums | Science Articles, Homework Help, Discussion**