Verifying the mean value theorem I got this question wrong on a test, I am not sure what went wrong.
Verify that the function satisfies the Mean Value Theorem and then find all numbers $c$ that satisfy the conclusion of the Mean Value Theorem.
$$f(x) = x + \sin 2x, [0, 2\pi]$$
This one I wasn't so sure what to do because I have no idea how to find two values that are equal to each other so I just plugged in $0$ and $2\pi$ and I got $-1$ as the answer which was wrong.
 A: We have to verify (copied from wikipedia...)

If a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a &lt b$, and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ such that $$f \ '(c) = \dfrac{f(b) - f(a)}{b - a}$$  

In this case, $f$ is continuous and differentiable, as it is the sum of terms that contain so-called $C^{\infty}$ functions. 
$a = 0; b = 2 \pi$
$f(a) = f(0) = 0$;
$f(b) = f(2 \pi) = 2 \pi$.  
Can you take the derivative and set it equal to $\dfrac{2 \pi - 0}{2 \pi - 0} = 1$ ?
A: Going by the language of your question, I think you're confusing between Mean Value Theorem and Rolle's Theorem. Let me state them here: 
Mean Value Theorem

If 
  
  
*
  
*a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a &lt b$, and 
  
*differentiable on the open interval $(a, b)$, 
  
  
  $\underline{\mbox{then}}$ there exists a point $c$ in $(a, b)$ such that $$f \ '(c) = \dfrac{f(b) - f(a)}{b - a}$$  

Rolle's Theorem

If 
  
  
*
  
*a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a &lt b$; 
  
*differentiable on the open interval $(a, b)$, and 
  
*$f(a)=f(b)$, 
  
  
  $\underline{\mbox{then}}$ there exists a point $c$ in $(a, b)$ such that $$f\ '(c) = 0$$

Let me write some more: 

  
*
  
*So, when you are asked to use Mean value theorem, you don't need to find values such that $f(\cdot_1)=f(\cdot_2)$. All you need to do is to verify that the continuity and differentiability hypotheses are true and proceed to find $c$ that is supposed to exist by MVT.  
  
*When you're asked to use Rolle's theorem, you need not find values such that $f(\cdot_1)=f(\cdot_2)$. All, you need to do is to check if the function agrees on the end points of the intervals already given to you and proceed to find $c$ asserted to exist by Rolle's Theorem. If the function does not agree on the end points, this function simply does not satisfy the hypothesis of Rolle's theorem and such a $c$ might not exist. 

A: $$
f(x) = x + \sin (2x)
$$
$$
\frac{f(2\pi) - f(0)}{2\pi-0} = \frac{(2\pi+\sin(2\cdot2\pi))- (0+\sin(2\cdot0))}{2\pi} = \frac{2\pi}{2\pi} = 1.
$$
$$
f'(x) = 1 + 2\cos(2x).
$$
So you need to show that there is a value $c$ strictly between $0$ and $2\pi$ such that $1+2\cos(2c)=1$.  That means you need $\cos(2c)=0$.  That happens if $2c$ is a right angle, i.e. $2c=\pi/2$, so $c=\pi/4$.
