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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.

3 Answers 3

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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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    @Jordan: It just so happens that the slope of the secant line **is** $1$ in this example!2012-04-18
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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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    For the most part, I'm an uninformed enthusiast!2012-04-17
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$ 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 $c=\pi/4$.