Mean value theorem section problem

[nhrock3]

f is continues in [0,1] and differentiable in (0,1)

f(0)=0 and forx\in(0,1) |f&#039;(x)|&lt;=|f(x)| and 0<a<1 prove:

(i)the set {|f(x)| : 0&lt;=x&lt;=a} has maximum

(ii)for every x\in(0,a] this innequality holds \frac{f(x)}{x}\leq max{|f(x)|:0&lt;=x&lt;=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 can't join them because its c there and not x

c is inside point x is on the border.



what to do?

 

[LCKurtz]

f is continues in [0,1] and differentiable in (0,1)

f(0)=0 and forx\in(0,1) |f&#039;(x)|&lt;=|f(x)| and 0<a<1 prove:

(i)the set {|f(x)| : 0&lt;=x&lt;=a} has maximum

(ii)for every x\in(0,a] this innequality holds \frac{f(x)}{x}\leq max{|f(x)|:0&lt;=x&lt;=a}

(iii)f(x)=0 for x\in[0,a]

(iiiן)f(x)=0 for x\in[0,1]

Use lower case in your tex tags as I have done above to make it readable. Also if you preview your posts you will see if it works. I will let others comment because I have a football game to watch