# How do you determine if rolles theorem can be applied to f(x)= 3sin(2x) on the interval [0, 2pi] and if so how do you find all the values of c in the interval for which f'(c)=0?

May 29, 2015

The Rolles theorem says that if:

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

then at least one $c \in \left(a , b\right)$ as if $f ' \left(c\right) = 0$ exists.

So:

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

To find $c$, we have to solve:

$y ' \left(c\right) = 0 \Rightarrow 6 \cos \left(2 c\right) = 0 \Rightarrow \cos \left(2 c\right) = 0 \Rightarrow$

$2 c = \frac{\pi}{2} + 2 k \pi \Rightarrow c = \frac{\pi}{4} + k \pi \Rightarrow$

${c}_{1} = \frac{\pi}{4}$

and

${c}_{2} = \frac{\pi}{4} + \pi = \frac{5}{4} \pi$.

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