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(C) The domain of the function $f(x) = \cos^{-1}x + 2\cot^{-1}x - 2x^3 - 4x$ is $[-1, 1]$.First,we find the derivative of $f(x)$:$f'(x) = \frac{-1}{\s

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$3\pi$

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(C) The domain of the function $f(x) = \cos^{-1}x + 2\cot^{-1}x - 2x^3 - 4x$ is $[-1, 1]$.First,we find the derivative of $f(x)$:$f'(x) = \frac{-1}{\sqrt{1-x^2}} - \frac{2}{1+x^2} - 6x^2 - 4$.Since $\frac{-1}{\sqrt{1-x^2}} Thus,$f(x)$ is a strictly decreasing function on $[-1, 1]$.The maximum value occurs at the left endpoint $x = -1$:$M = f(-1) = \cos^{-1}(-1) + 2\cot^{-1}(-1) - 2(-1)^3 - 4(-1) = \pi + 2(\frac{3\pi}{4}) + 2 + 4 = \pi + \frac{3\pi}{2} + 6 = \frac{5\pi}{2} + 6$.The minimum value occurs at the right endpoint $x = 1$:$m = f(1) = \cos^{-1}(1) + 2\cot^{-1}(1) - 2(1)^3 - 4(1) = 0 + 2(\frac{\pi}{4}) - 2 - 4 = \frac{\pi}{2} - 6$.The sum of the minimum and maximum values is:$m + M = (\frac{\pi}{2} - 6) + (\frac{5\pi}{2} + 6) = \frac{6\pi}{2} = 3\pi$.

The sum of the minimum and maximum values of the function $f(x) = \cos^{-1}x + 2\cot^{-1}x - 2x^3 - 4x$ is:

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