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
$$ f'(c)= \displaystyle{ f(\pi)-f(-\pi) \over \pi - (-\pi) } \ \ \ \ \longrightarrow $$
$$ 1 - 3 \sin c = \displaystyle{ (\pi + 3\cos \pi) - (-\pi + 3\cos (-\pi) ) \over 2\pi } \ \ \ \ \longrightarrow $$
$$ 1 - 3 \sin c = \displaystyle{ (\pi + 3(-1)) - (-\pi + 3(-1) ) \over 2\pi } \ \ \ \ \longrightarrow $$
$$ 1 - 3 \sin c = \displaystyle{ 2\pi \over 2\pi } \ \ \ \ \longrightarrow $$
$$ 1 - 3 \sin c = 1 \ \ \ \ \longrightarrow $$
$$ - 3 \sin c = 0 \ \ \ \ \longrightarrow $$
$$ \sin c = 0 \ \ \ \ \longrightarrow $$
$$ c = 0 $$
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