I just took a test and I could not figure out this problem.
I was suppose to show that the function satisfies the three conditions for the mean value theorem and then use it.
$x+\sin(2x)$ for $x \in [0,2\pi]$
I forgot what $\sin(2x)$ means, if it is $2\sin x$ or just $\sin 2x$ or whatever so I had no idea how to find zeroes or any function that equals another on this. I entered the function into a table and looked for values at ever .10 numbers and I couldn't find anything.
I do know that the function does go up and down and there there is at least one derivative on 0-2pi. I do not even know how to start this problem.
Best Answer
The Mean Value Theorem states:
Our function here is $f(x) = x+\sin(2x)$, and $[a,b]$ is $[0,2\pi]$.
The function is continuous everywhere: $x$ is continuous everywhere, so $2x$ is continuous everywhere. Since $\sin(u)$ is continuous everywhere, the composition $\sin(2x)$ is continuous everywhere. Since $x$ and $\sin(2x)$ are both continuous everywhere, their sum is continuous everywhere. And since $f(x) = x+\sin(2x)$ is continuous everywhere, it is in particular continuous on $[0,2\pi]$.
Similarly, each of the functions mentioned is differentiable everywhere, so $f(x)=x+\sin(2x)$ is differentiable everywhere. Since it is differentiable everywhere, it is also differentiable on $(0,2\pi)$.
According to the Mean Value Theorem, there must exist at least one point $c$ in $(0,2\pi)$ where $$\begin{align*} f'(c) &= \frac{f(2\pi)-f(0)}{2\pi - 0}\\ &= \frac{\Bigl( 2\pi + \sin(2(2\pi))\Bigr) - \Bigl( 0 + \sin(2(0))\Bigr)}{2\pi}\\ &=\frac{2\pi + \sin(4\pi) - 0 - \sin (0)}{2\pi}\\ &= \frac{2\pi + 0 - 0 + 0}{2\pi}\\ &=\frac{2\pi}{2\pi}\\ &= 1. \end{align*}$$
So the Mean Value Theorem tells us that there is at least one point $c$ in $(0,2\pi)$ where $f'(c) = 1$.
That would be it.
You may also want to verify that the conclusion is indeed true by exhibiting a point $c$ in $(0,2\pi)$ where this is true. We have $$f'(x) = 1 + \cos(2x)(2x)' = 1 + 2\cos(2x).$$ So $f'(c) = 1$ if and only if $1+2\cos(2c) = 1$, if and only if $2\cos(2c)=0$, if and only if $\cos(2c)=0$.
Cosine is $0$ on the odd multiples of $\frac{\pi}{2}$; the values of $c$ on $(0,2\pi)$ where $\cos(2c)=0$ are $c=\frac{\pi}{4}$, $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{4}$. So indeed, there is at least one (in fact four) points $c$ in $(0,2\pi)$ where $f'(c) = \frac{f(b)-f(a)}{b-a}$, as predicted by the Mean Value Theorem.