Let f(x) = x + sin2x on [0, 2(pi)] find two numbers c that satisfy the conclusion of hte Mean Value Theorem. (note there are four such numbers.)

2. $a=0, \;\ b=2\pi$

$f(x)=x+sin(2x)$

$f(a)=0, \;\ f(b)=2\pi$

$f'(x)=2cos(2x)+1$

$f'(c)=2cos(2c)+1$

Also, $f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{2\pi-0}{2\pi-0}=1$

So, we have $2cos(2c)+1=1$

Solving for c we find $c=\frac{(2c-1){\pi}}{4}$

When c= 1 through 4, we get:

$c=\frac{\pi}{4}, \;\ \frac{3\pi}{4}, \;\ \frac{5\pi}{4}, \;\ \frac{7\pi}{4}$

are the four points which satisfy the MVT over the given interval.