# 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.

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We have to verify (copied from wikipedia...)

If a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a < b$, 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 $\dfrac{2 \pi - 0}{2 \pi - 0} = 1$ ?

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I disagree. The question said only that one should verify that the conclusion of the mean value theorem is true in this instance. For that you don't need to check that the mean value theorem is actually applicable. You might have, for example, a discontinuous function, so the theorem is not applicable at all, but in some such cases the conclusion will still hold. –  Michael Hardy Apr 17 '12 at 1:40
@MichaelHardy: I agree with your explanation ("For that... will still hold."), but not with your interpretation that we are considering the conclusion of the MVT rather than the hypothesis. Maybe this is a philosophical difference... ??? –  The Chaz 2.0 Apr 17 '12 at 1:47
Alright, maybe "satisfies" is open to interpretation..... –  Michael Hardy Apr 17 '12 at 2:04
@TheChaz What does c to the infinity mean? –  user138246 Apr 17 '12 at 3:11
I mean smooth functions. I know that's a bit over your head at the moment... it means that you can take derivatives forever (more or less...). –  The Chaz 2.0 Apr 17 '12 at 3:18

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

1. a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a < b$, and
2. 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

1. a function $f(x)$ is continuous on the closed interval $[a, b]$, where $a < b$;
2. differentiable on the open interval $(a, b)$, and
3. $f(a)=f(b)$,

$\underline{\mbox{then}}$ 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.

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The theorem statements were copied from @TheChaz 's answer and modified. :-) –  user21436 Apr 17 '12 at 1:33
For the MVT how will I know that the function doesn't just go straight up or something? Wouldn't I have to find an f(a) that equals an f(b)? –  user138246 Apr 17 '12 at 3:14
This would be a great addition to a Calc I "Overview" answer. (Referring to this meta post ) –  The Chaz 2.0 Apr 17 '12 at 3:22
@TheChaz I agree; but if that will materialize, I am willing to substantiate this post with more information; work out a few examples and the likes. May be a Sub Wiki with links to SE would be a good project. For group Theory, Vipul of UChicago has a Sub Wiki which links to some posts here. (Sorry if I am telling you what you already know.) –  user21436 Apr 17 '12 at 5:04
For the most part, I'm an uninformed enthusiast! –  The Chaz 2.0 Apr 17 '12 at 5:12

$$f(x) = x + \sin (2x)$$ $$\frac{f(2\pi) - f(0)}{2\pi-0} = \frac{(2\pi+\sin(2\cdot2\pi))- (0+\sin(2\cdot0))}{2\pi} = \frac{2\pi}{2\pi} = 1.$$ $$f'(x) = 1 + 2\cos(2x).$$ 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 $c=\pi/4$.

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