By MVT, for some . Now, by assumption, and since .
Hint: use \{ and \} to make curly braces show in LaTeX.
f is continues in [0,1] and differentiable in (0,1)
f(0)=0 and for and 0<a<1 prove:
(i)the set has maximum
(ii)for every this innequality holds
(iii)f(x)=0 for
(iiiן)f(x)=0 for
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?