Mean Value Theorem Solution 5

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SOLUTION 5: We are given the function

$f(x) = \sin 2x$ and the interval

$[0, \pi]$ . This function is continuous on the closed interval

$[0, \pi]$ since it is the functional composition of continuous functions

$y=2x$ (polynomial) and

$y= \sin x$ (well-known continuous). The derivative of

$f$ is

$f'(x) = 2 \cos 2x$ We can now see that

$f$ is differentiable on the open interval

$(0, \pi)$ . The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of

$c$ in the open interval

$(0, \pi)$ . Then

$f'(c)= \displaystyle{ f(\pi)-f(0) \over \pi - 0 } \ \ \ \ \longrightarrow$

$2 \cos 2c = \displaystyle{ \sin 2(\pi) - \sin 2(0) \over \pi } \ \ \ \ \longrightarrow$

$2 \cos 2c = \displaystyle{ \sin 2\pi - \sin 0 \over \pi } \ \ \ \ \longrightarrow$

$2 \cos 2c = \displaystyle{ 0-0 \over \pi } = 0 \ \ \ \ \longrightarrow$

$\cos 2c = 0 \ \ \ \ \longrightarrow$

$2c = \displaystyle{ \pi \over 2 } \ \ or \ \ \displaystyle{ 3\pi \over 2 } \ \ \ \ \longrightarrow$

$c = \displaystyle{ \pi \over 4 } \ \ or \ \ \displaystyle{ 3\pi \over 4 }$

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