SOLUTION 6: We are given the function $f(x) = x + 3\cos x$ and the interval $[-\pi, \pi]$. This function is continuous on the closed interval $[0, \pi]$ since it is the sum of continuous functions $y=x$ (polynomial) and $y= 3 \cos x$ (well-known continuous). The derivative of $f$ is $$f'(x) = 1 + 3(- \sin x) = 1 - 3 \sin x$$ We can now see that $f$ is differentiable on the open interval $(-\pi, \pi)$. 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 $(-\pi, \pi)$. Then $$f'(c)= \displaystyle{ f(\pi)-f(-\pi) \over \pi - (-\pi) } \ \ \ \ \longrightarrow$$ $$1 - 3 \sin c = \displaystyle{ (\pi + 3\cos \pi) - (-\pi + 3\cos (-\pi) ) \over 2\pi } \ \ \ \ \longrightarrow$$ $$1 - 3 \sin c = \displaystyle{ (\pi + 3(-1)) - (-\pi + 3(-1) ) \over 2\pi } \ \ \ \ \longrightarrow$$ $$1 - 3 \sin c = \displaystyle{ 2\pi \over 2\pi } \ \ \ \ \longrightarrow$$ $$1 - 3 \sin c = 1 \ \ \ \ \longrightarrow$$ $$- 3 \sin c = 0 \ \ \ \ \longrightarrow$$ $$\sin c = 0 \ \ \ \ \longrightarrow$$ $$c = 0$$