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 $$f'(c)= \displaystyle{ f(\pi)-f(0) \over \pi - 0 } \ \ \ \ \longrightarrow$$ $$2 \cos 2c = \displaystyle{ \sin 2(\pi) - \sin 2(0) \over \pi } \ \ \ \ \longrightarrow$$ $$2 \cos 2c = \displaystyle{ \sin 2\pi - \sin 0 \over \pi } \ \ \ \ \longrightarrow$$ $$2 \cos 2c = \displaystyle{ 0-0 \over \pi } = 0 \ \ \ \ \longrightarrow$$ $$\cos 2c = 0 \ \ \ \ \longrightarrow$$ $$2c = \displaystyle{ \pi \over 2 } \ \ or \ \ \displaystyle{ 3\pi \over 2 } \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ \pi \over 4 } \ \ or \ \ \displaystyle{ 3\pi \over 4 }$$