Morgan is asked to find the sixth degree Taylor polynomial centered at x = 0 for 12 -x* and provides the following argument. k +1 k=0 f (x) = g(x) cos(x), where g(x) = Since cos() = -1)* (2k)! -x2k, we can multiply the Taylor series. COS k=0 (-1)* (2k)! 12(-1)* (k + 1)(2k)!" 12 f(x) = k +1 k=0 k=0 Thus, the sixth degree Taylor polynomial is 12 – 3x° + 6 Determine if Morgan's argument is correct or incorrect. If the argument is incorrect, explain what error or errors Morgan made, then calculate the requested sixth degree Taylor polynomial correctly.

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