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The Rolles theorem says that if:

  1. $y=f(x)$ is a continue function in a set $[a,b]$;
  2. $y=f(x)$ is a derivable function in a set $(a,b)$;
  3. $f(a)=f(b)$;

then at least one $cin(a,b)$ as if $f'(c)=0$ exists.

So:

  1. $y=3sin(2x)$ is a function that is continue in all $RR$, and so it is in $[0,2pi]$;
  2. $y'=6cos(2x)$ is a function continue in all $RR$, so our function is derivable in all $RR$, so it is in $[0,2pi]$;
  3. $f(0)=f(2pi)=0$.

To find $c$, we have to solve:

$y'(c)=0rArr6cos(2c)=0rArrcos(2c)=0rArr$

$2c=pi/2+2kpirArrc=pi/4+kpirArr$

$c_1=pi/4$

and

$c_2=pi/4+pi=5/4pi$.

(both the values are $in[0,2pi]$).

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