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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