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(B) Given $f(x) = 2 \log_e(x-2) - x^2 + 4x + 1$.For the function to be increasing,we must have $f'(x) > 0$.$f'(x) = \frac{2}{x-2} - 2x + 4$.$f'(x) = \

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$(2, 3)$

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(B) Given $f(x) = 2 \log_e(x-2) - x^2 + 4x + 1$.For the function to be increasing,we must have $f'(x) > 0$.$f'(x) = \frac{2}{x-2} - 2x + 4$.$f'(x) = \frac{2 - 2x(x-2) + 4(x-2)}{x-2} = \frac{2 - 2x^2 + 4x + 4x - 8}{x-2} = \frac{-2x^2 + 8x - 6}{x-2}$.$f'(x) = \frac{-2(x^2 - 4x + 3)}{x-2} = \frac{-2(x-3)(x-1)}{x-2}$.For $f'(x) > 0$,we need $\frac{(x-3)(x-1)}{x-2} Using the wavy curve method,the intervals are $(-\infty, 1) \cup (2, 3)$.Since the domain of $\log_e(x-2)$ is $x > 2$,we intersect the intervals with $x > 2$.The resulting set is $(2, 3)$.

The complete set of values of $x$ for which the function $f(x) = 2 \log_e(x-2) - x^2 + 4x + 1$ is increasing is:

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