# Mean value theorem with trig functions

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.

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Slow down and tell us exactly what the problem was. Were you supposed to find a $c\in[0,2\pi]$ satisfying the conclusion of the mean value theorem, i.e., $f'(c)=\frac{f(2\pi)-f(0)}{2\pi}$? – Brian M. Scott Apr 10 '12 at 1:17
I edited the post, but I have forgotten what the question was asking me exactly. Also I am not sure what Brian is asking. – user138246 Apr 10 '12 at 1:20
If there is no question, what answers to do you expect? – Aryabhata Apr 10 '12 at 1:21
For the record, if you're referring to the double angle identity for $\sin(2x)$, it is $\sin(2x)$ = $2 \sin x \cos x$. – Joe Apr 10 '12 at 1:25
Is sin2x the same as 2sinx? – user138246 Apr 10 '12 at 1:48

The Mean Value Theorem states:

If $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least on $c\in (a,b)$ such that $$f'(c) = \frac{f(b)-f(a)}{2\pi}.$$

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.

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I got 1.117 or something for the answer, so I definitely failed this test. Anyways I originally had 1 as the answer but I realized that it does not satisfy the MVT because f(0) is not equal to f(2pi) – user138246 Apr 10 '12 at 1:34
@Jordan: MVT does not require $f(0)=f(2\pi)$. That's Rolle's Theorem. I've stated exactly what the MVT says in the greybox. I have to guess, because I don't know what the xact wording of the problem was, but I suspect you were supposed to verify the hypothesis (that is, explain or show that $f(x)=x+\sin(2x)$ is continuous on the closed interval, and differentiable on the open interval), and then "apply it" to say what is the specific conclusion you get for this particular function in this particular interval (that there is a $c$ in $(0,2\pi)$ with $f'(c) = (f(b)-f(a))/(b-a) = 1$. (cont) – Arturo Magidin Apr 10 '12 at 1:37
(cont) Whether or not you were also supposed to explicitly exhibit some $c$ with this property, I do not know. – Arturo Magidin Apr 10 '12 at 1:37